Indefinite Integration

$$\begin{gathered} \int x^2·\frac{\left(x{\sec }^{2}x+\tan x\right)}{(x\tan x+1{)}^2}dx \\ I II \end{gathered}$$

Using integral by parts

$$\begin{gathered} =\frac{-x^2}{(x\tan x+1)}+\int \frac{2x}{x\tan x+1}dx \\ let I=\int \frac{2x}{x\tan x+1}dx \end{gathered}$$

$I=2\int \frac{x}{x\tan x+1}dx$

$$\begin{gathered} =2\int \frac{x\cos x}{x\sin x+\cos x}dx \\ \text{ Let }x\sin x+\cos x=t \\ (x\cos x+\sin x-\sin x)dx=dt \end{gathered}$$

$$\begin{gathered} =2\int \frac{dt}{t}=2\log t+c^{\text{'}} \\ =2\log |x\sin x+\cos x|+c^{\text{'}} \\ \therefore \int \frac{x^2\left(x{\sec }^{2}x+\tan x\right)}{(x\tan x+1{)}^2}dx \\ =\frac{-x^2}{x\tan x+1}+2\log |x\sin x+\cos x|+c \end{gathered}$$

$\int \frac{2-\tan x}{3+\tan x}dx=\int \frac{2\cos x-\sin x}{3\cos x+\sin x}dx$

$2\cos x-\sin x=A(3\cos x+\sin x)+B(\cos x-3\sin x)$

$3A + B = 2$

$A – 3B = –1$

$\Rightarrow \mathrm{A}=\frac{1}{2}, \mathrm{B}=\frac{1}{2}$

$\therefore \int \frac{2\cos x-\sin x}{3\cos x+\sin x}dx$

\(=\frac{1}{2}\left[\int \frac{-3 \sin x+\cos x}{3 \cos x+\sin x} d x+\int \frac{3 \cos x+\sin x}{3 \cos x+\sin x} d x\right]\)

$=\frac{x}{2}+\frac{1}{2}\ln |3\cos x+\sin x|+C$

$=\frac{1}{2}(x+\ln |3\cos x+\sin x\left|\right)+C$

$=\frac{1}{2}(\alpha x+\ln |\beta \sin x+\gamma \cos x\left|\right)+C$

$\alpha =1,\beta =1,\gamma =3$

$\therefore \alpha +\frac{\gamma }{\beta }=1+\frac{3}{1}=4$

$\int \frac{2-\tan x}{3+\tan x}dx=\int \frac{2\cos x-\sin x}{3\cos x+\sin x}dx$

$2\cos x-\sin x=A(3\cos x+\sin x)+B(\cos x-3\sin x)$

$3A + B = 2$

$A – 3B = –1$

$\Rightarrow \mathrm{A}=\frac{1}{2}, \mathrm{B}=\frac{1}{2}$

$\therefore \int \frac{2\cos x-\sin x}{3\cos x+\sin x}dx$

\(=\frac{1}{2}\left[\int \frac{-3 \sin x+\cos x}{3 \cos x+\sin x} d x+\int \frac{3 \cos x+\sin x}{3 \cos x+\sin x} d x\right]\)

$=\frac{x}{2}+\frac{1}{2}\ln |3\cos x+\sin x|+C$

$=\frac{1}{2}(x+\ln |3\cos x+\sin x\left|\right)+C$

$=\frac{1}{2}(\alpha x+\ln |\beta \sin x+\gamma \cos x\left|\right)+C$

$\alpha =1,\beta =1,\gamma =3$

$\therefore \alpha +\frac{\gamma }{\beta }=1+\frac{3}{1}=4$

$$\begin{gathered} \text{Using Integration by parts} \\ \int x^3\sin xdx=-x^3\cos x+\int 3x^2\cos xdx \\ =-x^3\cos x+3x^2\sin x-\int 6x\sin xdx \\ \text{ So }g\left(x\right)=-x^3\cos x+3x^2\sin x \\ +6x\cos x-6\sin x \\ g\left(\frac{\pi }{2}\right)=\frac{3{\pi }^2}{4}-6 \\ {g}^{\text{'}}\left(x\right)=-3x^2\cos x+x^3\sin x \\ +3x^2\cos x+6\cos x-6\cos x \\ {g}^{\text{'}}\left(\frac{\pi }{2}\right)=\frac{{\pi }^3}{8} \\ 8\left(g\left(\frac{\pi }{2}\right)+g^{\text{'}}\left(\frac{\pi }{2}\right)\right)={\pi }^3+6{\pi }^2-48 \\ \text{ So }\alpha +\beta -\gamma =55 \end{gathered}$$

$$\begin{gathered} \text{let} x=t^4 \\ dx=4t^{3}dt \\ \text{then} \int \frac{1}{x^{\frac{1}{4}}\left(1+x^{\frac{1}{4}}\right)}dx \\ =\int \frac{4t^{3}dt}{t(1+t)} \\ =\int \frac{4{\mathrm{t}}^2}{1+\mathrm{t}}\mathrm{dt} \\ =4\int \frac{\left({\mathrm{t}}^2-1\right)+1}{1+\mathrm{t}}\mathrm{dt} \\ =4\int \left(\mathrm{t}-1\right)+\frac{1}{\mathrm{t}+1}\mathrm{dt} \\ =4\left\{\frac{(\mathrm{t}-1{)}^2}{2}+\ln \left(t+1\right)\right\}+\mathrm{c} \\ \mathrm{hence} \mathrm{f}\left(\mathrm{x}\right)=2{\left({\mathrm{x}}^{\frac{1}{4}}-1\right)}^2+4\ln \left(1+{\mathrm{x}}^{\frac{1}{4}}\right)+\mathrm{c} \\ \mathrm{f}\left(0\right)=-6\Rightarrow 2+4\mathrm{\ell n}\left(1\right)+C=-6 \\ \mathrm{C}=-8 \\ \mathrm{now} \\ \mathrm{f}\left(1\right)=4\ell n2-8 \\ =4\left(\mathrm{\ell n}2-2\right) \end{gathered}$$

$f\left(x\right)=\int \left(\frac{{\displaystyle \frac{7}{x^8}+\frac{9}{x^{10}}}}{{\left(\frac{1}{x^9}+\frac{1}{x^7}+2\right)}^2}\right)dx$

Put $t=\frac{1}{x^9}+\frac{1}{x^7}+2$

$\Rightarrow \frac{dt}{dx}=\frac{-9}{x^{10}}-\frac{7}{x^8}$

$f\left(x\right)=\int \frac{-dt}{t^2}=\frac{1}{t}+C$

$f\left(x\right)=\frac{1}{\frac{1}{x^9}+\frac{1}{x^7}+2}+C$

$=\frac{x^9}{1+x^2+2x^9}+C$

Given $f\left(1\right)=\frac{1}{4}=\frac{1}{4}+C\Rightarrow C=0$

$f\left(x\right)=\frac{x^9}{1+x^2+2x^9}$

$f^{\text{'}}\left(x\right)=\frac{\left(1+x^2+2x^9\right)9x^8-x^9\left(2x+18x^8\right)}{{\left(1+x^2+2x^9\right)}^2}$

$f^{\text{'}}\left(1\right)=\frac{36-20}{16}=1$

\(\mathrm{B}=\operatorname{adj}(\operatorname{adj} \mathrm{A})\)

\(|\mathrm{B}|=81=|\mathrm{A}|^4 \Rightarrow|\mathrm{~A}|=\pm 3\)

$A=\left[\begin{array}{ccc}0& 0& 1\\ \frac{1}{4}& 1& 1\\ {\alpha }^2& 4& 1\end{array}\right]$

\(|\mathrm{A}|=\left(1-\alpha^2\right)\)

\(1-\alpha^2=3,-3\)

\(\Rightarrow \alpha^2=-2,4\)

value of ${\alpha }^2=4$

Let $4x+6=\frac{1}{t}\Rightarrow x=\frac{\frac{1}{t}-6}{4}$

$4dx=-\frac{dt}{t^2}$

$\int \frac{3dx}{(4x+6)\sqrt{4{(x+1)}^2-1}}$

$=\int \frac{3(-dt)}{4t^2\times \frac{1}{t}\sqrt{4{\left(\frac{\frac{1}{t}-2}{4}\right)}^2-1}}$

$=-\frac{3}{4}\int \frac{dt}{t\sqrt{\frac{{(1-2t)}^2}{4t^2}-1}}$

$=-\frac{3}{4}\int \frac{dt\left(2t\right)}{t\sqrt{1-4t}}$

$=-\frac{3}{2}\int \frac{dt}{\sqrt{1-4t}}=-\frac{3}{2}\left(\frac{\sqrt{1-4t}}{\frac{1}{2}\times -4}\right)+c$

$=\frac{3}{4}\sqrt{1-4t}+c$ $\left(\because t=\frac{1}{4x+6}\right)$

$=\frac{3}{4}\sqrt{1-4\left(\frac{1}{4x+6}\right)}+c$

$=\frac{3}{4}\sqrt{\frac{4x+6-4}{4x+6}}+c$

$I\left(x\right)=\frac{3}{4}\sqrt{\frac{4x+2}{4x+6}}+c$

Now, $I\left(0\right)=\frac{3}{4}\sqrt{\frac{2}{6}}+c$

$$\begin{gathered} I\left(0\right)=\frac{\sqrt{3}}{4}+c=\frac{\sqrt{3}}{4}+20 \left(given\right) \\ \Rightarrow c=20 \end{gathered}$$

Hence, $I\left(x\right)=\frac{3}{4}\sqrt{\frac{4x+2}{4x+6}}+20$

$I\left(\frac{1}{2}\right)=\frac{3}{4}\sqrt{\frac{4}{8}}+20=\frac{3}{4\sqrt{2}}+20=\frac{3\sqrt{2}}{8}+20$

Thus, $a+b+c=3+8+20=31$

$$\begin{gathered} \mathrm{I}\left(\mathrm{x}\right)=\int \frac{\mathrm{dx}}{(\mathrm{x}-11{)}^{\frac{11}{13}}(\mathrm{x}+15{)}^{\frac{15}{13}}}. \\ \mathrm{I}\left(\mathrm{x}\right)=\int \frac{\mathrm{dx}}{\left(\frac{x-11}{x+15}{)}^{\frac{11}{13}}\right(\mathrm{x}+15{)}^2}. \\ \text{ Let }\frac{\mathrm{x}-11}{\mathrm{x}+15}=\mathrm{t}\Rightarrow \frac{26}{(\mathrm{x}+15{)}^2}\mathrm{dx}=\mathrm{dt} \\ I\left(\mathrm{x}\right)=\frac{1}{26}\int \frac{\mathrm{dt}}{{\mathrm{t}}^{11/13}}=\frac{1}{26}·\frac{{\mathrm{t}}^{2/13}}{2/13}+C \\ I\left(\mathrm{x}\right)=\frac{1}{4}{\left(\frac{x-11}{x+15}\right)}^{2/13}+C \\ \text{Now,} \\ \mathrm{I}\left(37\right)-\mathrm{I}\left(24\right)=\frac{1}{4}{\left(\frac{26}{52}\right)}^{2/13}-\frac{1}{4}{\left(\frac{13}{39}\right)}^{2/13} \\ =\frac{1}{4}\left(\frac{1}{2^{2/13}}-\frac{1}{3^{2/13}}\right) \\ \mathrm{I}\left(37\right)-\mathrm{I}\left(24\right)=\frac{1}{4}\left(\frac{1}{4^{1/13}}-\frac{1}{9^{1/13}}\right) \\ \text{Compare with,}\mathrm{I}\left(37\right)-\mathrm{I}\left(24\right)=\frac{1}{4}\left(\frac{1}{b^{\frac{1}{13}}}-\frac{1}{c^{\frac{1}{13}}}\right) \\ \text{we get} b=4 \& c=9 \\ \text{Now,} 3\left(b+c\right)=3\left(4+9\right)=39 \end{gathered}$$

$\int \frac{2-\tan x}{3+\tan x}dx=\int \frac{2\cos x-\sin x}{3\cos x+\sin x}dx$

$2\cos x-\sin x=A(3\cos x+\sin x)+B(\cos x-3\sin x)$

$3A + B = 2$

$A – 3B = –1$

$\Rightarrow \mathrm{A}=\frac{1}{2}, \mathrm{B}=\frac{1}{2}$

$\therefore \int \frac{2\cos x-\sin x}{3\cos x+\sin x}dx$

$=\frac{x}{2}+\frac{1}{2}\ln |3\cos x+\sin x|+C$

$=\frac{1}{2}(x+\ln |3\cos x+\sin x\left|\right)+C$

$=\frac{1}{2}(\alpha x+\ln |\beta \sin x+\gamma \cos x\left|\right)+C$

$\alpha =1,\beta =1,\gamma =3$

$\therefore \alpha +\frac{\gamma }{\beta }=1+\frac{3}{1}=4$

$$\begin{gathered} I={\int }_0^{\frac{\pi }{2}}\frac{\sin \left(\frac{\pi }{4}+x\right)+\sin \left(\frac{3\pi }{4}+x\right)}{\cos x+\sin x}dx \\ I={\int }_0^{\frac{\pi }{2}}\frac{2\sin {\displaystyle \frac{\mathrm{\pi }}{4}}.\cos x}{\sin x+\cos x}dx \\ I=\sqrt{2}{\int }_0^{\frac{\pi }{2}}\frac{\cos x}{\sin x + \cos x}dx ......\left(i\right) \\ I=\sqrt{2}{\int }_0^{\frac{\pi }{2}}\frac{\sin x}{\sin x + \cos x}dx ......\left(ii\right) \end{gathered}$$

$$\begin{gathered} \text{Adding eqn }\left(i\right) \& \left(ii\right), \text{we get} \\ 2I=\sqrt{2}{\int }_0^{\frac{\pi }{2}}\frac{\sin x + \cos x}{\sin x + \cos x}dx \\ 2I = \sqrt{2}{\int }_0^{\frac{\pi }{2}}1dx \\ 2I = \sqrt{2}.\frac{\mathrm{\pi }}{2} \\ I = \frac{\mathrm{\pi }}{2\sqrt{2}} \end{gathered}$$

\(\begin{aligned}
& \text { Let: } f(x)=\frac{x}{\sqrt{1-x^2}} \sin ^{-1} x \\
& f^{\prime}(x)=\sin ^{-1} x\left(\frac{\sqrt{1-x^2}-\frac{x(-2 x)}{2 \sqrt{1-x^2}}}{(1-x)^2}\right)+\frac{x}{\sqrt{1-x^2}} \times \frac{1}{\sqrt{1-x^2}} \\
& f^{\prime}(x)=\sin ^{-1} \times\left(\frac{1-x^2+x^2}{\left(1-x^2\right)^{3 / 2}}\right)+\frac{x}{1-x^2} \\
& g(x)=e^x \cdot \frac{x \sin ^{-1} x}{\sqrt{1-x^2}} \\
& g\left(\frac{1}{2}\right)=\frac{\sqrt{e} \times \frac{1}{2} \times \frac{\pi}{6}}{\frac{\sqrt{3}}{2}}=\frac{\pi}{6} \sqrt{\frac{e}{3}}
\end{aligned}\)

$$\begin{gathered} \mathrm{I}\left(\mathrm{x}\right)=\int \frac{\mathrm{dx}}{(\mathrm{x}-11{)}^{\frac{11}{13}}(\mathrm{x}+15{)}^{\frac{15}{13}}}. \\ \mathrm{I}\left(\mathrm{x}\right)=\int \frac{\mathrm{dx}}{\left(\frac{x-11}{x+15}{)}^{\frac{11}{13}}\right(\mathrm{x}+15{)}^2}. \\ \text{ Let }\frac{\mathrm{x}-11}{\mathrm{x}+15}=\mathrm{t}\Rightarrow \frac{26}{(\mathrm{x}+15{)}^2}\mathrm{dx}=\mathrm{dt} \\ I\left(\mathrm{x}\right)=\frac{1}{26}\int \frac{\mathrm{dt}}{{\mathrm{t}}^{11/13}}=\frac{1}{26}·\frac{{\mathrm{t}}^{2/13}}{2/13}+C \\ I\left(\mathrm{x}\right)=\frac{1}{4}{\left(\frac{x-11}{x+15}\right)}^{2/13}+C \\ \text{Now,} \\ \mathrm{I}\left(37\right)-\mathrm{I}\left(24\right)=\frac{1}{4}{\left(\frac{26}{52}\right)}^{2/13}-\frac{1}{4}{\left(\frac{13}{39}\right)}^{2/13} \\ =\frac{1}{4}\left(\frac{1}{2^{2/13}}-\frac{1}{3^{2/13}}\right) \\ \mathrm{I}\left(37\right)-\mathrm{I}\left(24\right)=\frac{1}{4}\left(\frac{1}{4^{1/13}}-\frac{1}{9^{1/13}}\right) \\ \text{Compare with,}\mathrm{I}\left(37\right)-\mathrm{I}\left(24\right)=\frac{1}{4}\left(\frac{1}{b^{\frac{1}{13}}}-\frac{1}{c^{\frac{1}{13}}}\right) \\ \text{we get} b=4 \& c=9 \\ \text{Now,} 3\left(b+c\right)=3\left(4+9\right)=39 \end{gathered}$$

$$\begin{gathered} \int x^2·\frac{\left(x{\sec }^{2}x+\tan x\right)}{(x\tan x+1{)}^2}dx \\ I II \end{gathered}$$

Using integral by parts

$$\begin{gathered} =\frac{-x^2}{(x\tan x+1)}+\int \frac{2x}{x\tan x+1}dx \\ let I=\int \frac{2x}{x\tan x+1}dx \end{gathered}$$

$I=2\int \frac{x}{x\tan x+1}dx$

$$\begin{gathered} =2\int \frac{x\cos x}{x\sin x+\cos x}dx \\ \text{ Let }x\sin x+\cos x=t \\ (x\cos x+\sin x-\sin x)dx=dt \end{gathered}$$

$$\begin{gathered} =2\int \frac{dt}{t}=2\log t+c^{\text{'}} \\ =2\log |x\sin x+\cos x|+c^{\text{'}} \\ \therefore \int \frac{x^2\left(x{\sec }^{2}x+\tan x\right)}{(x\tan x+1{)}^2}dx \\ =\frac{-x^2}{x\tan x+1}+2\log |x\sin x+\cos x|+c \end{gathered}$$

\(f^{\prime \prime}(x)=g^{\prime \prime}(x)\)

By integration:

\(f^{\prime}(x)=g^{\prime}(x)+c \)
Put \(x=1 \)
\( 4=2+c \)
\( c=2\)
\( f^{\prime}(x)=g^{\prime}(x)+2 \)
\( f(x)=g(x)+2 x+K \)
Put \( x=2 \)
\( f(2)=g(2)+4+K\)
\( 3=9+4+K \)
\( K=-10 \)
\( f(x)=g(x)+2 x-10\)
Put \(\mathrm{x}=25\)

\(f(25)-g(25)=50-10=40\)

$\int \left(\frac{1-5{\cos }^{2}x}{{\sin }^{5}x{\cos }^{2}x}\right)dx$

$=\int \frac{dx}{{\sin }^5{\cos }^{2}x}-5\int \frac{dx}{{\sin }^{5}x}$

$=\int \frac{{\sec }^{2}xdx}{{\sin }^{5}x}-5\int \frac{dx}{{\sin }^{5}x}$

Using Integration by parts in first integral

$=\frac{\tan x}{{\sin }^{5}x}-\int \left(-\frac{5}{{\sin }^{6}x}\right)\cdot \cos x\cdot \tan xdx-5\int \frac{dx}{{\sin }^{5}x}$

$=\frac{\text{tan}x}{{\text{sin}}^{5}x}+c$

$f\left(x\right)=\frac{\tan x}{{\sin }^{5}x}$

$f\left(\frac{\pi }{6}\right)-f\left(\frac{\pi }{4}\right)=\frac{2^5}{\sqrt{3}}-{\left(\sqrt{2}\right)}^5$

$=\frac{32}{\sqrt{3}}-4\sqrt{2}$

$=\frac{4}{\sqrt{3}}\left(8-\sqrt{6}\right)$

\(\begin{aligned}
& \int \frac{d x}{(x-1)^2\left(\frac{x+5}{x-1}\right)^{15 / 13}} \\
& \frac{x+5}{x-1}=t \\
& \frac{-16}{(x-1)^2} d x=d t \\
& -\frac{1}{16} \int \frac{d t}{t^{15 / 13}} \\
& -\frac{1}{16} \frac{t^{-15 / 13+1}}{-15 / 13+1}+C \\
& \frac{-1}{16} \frac{t^{-2 / 13}}{-2 / 13}+C \\
& \frac{13}{32}\left(\frac{x+15}{x-1}\right)^{-2 / 13}+C
\end{aligned}\)

\(\begin{aligned}
& \text { Let: } f(x)=\frac{x}{\sqrt{1-x^2}} \sin ^{-1} x \\
& f^{\prime}(x)=\sin ^{-1} x\left(\frac{\sqrt{1-x^2}-\frac{x(-2 x)}{2 \sqrt{1-x^2}}}{(1-x)^2}\right)+\frac{x}{\sqrt{1-x^2}} \times \frac{1}{\sqrt{1-x^2}} \\
& f^{\prime}(x)=\sin ^{-1} \times\left(\frac{1-x^2+x^2}{\left(1-x^2\right)^{3 / 2}}\right)+\frac{x}{1-x^2} \\
& g(x)=e^x \cdot \frac{x \sin ^{-1} x}{\sqrt{1-x^2}} \\
& g\left(\frac{1}{2}\right)=\frac{\sqrt{e} \times \frac{1}{2} \times \frac{\pi}{6}}{\frac{\sqrt{3}}{2}}=\frac{\pi}{6} \sqrt{\frac{e}{3}}
\end{aligned}\)

$$\begin{gathered} \text{let} x=t^4 \\ dx=4t^{3}dt \\ \text{then} \int \frac{1}{x^{\frac{1}{4}}\left(1+x^{\frac{1}{4}}\right)}dx \\ =\int \frac{4t^{3}dt}{t(1+t)} \\ =\int \frac{4{\mathrm{t}}^2}{1+\mathrm{t}}\mathrm{dt} \\ =4\int \frac{\left({\mathrm{t}}^2-1\right)+1}{1+\mathrm{t}}\mathrm{dt} \\ =4\int \left(\mathrm{t}-1\right)+\frac{1}{\mathrm{t}+1}\mathrm{dt} \\ =4\left\{\frac{(\mathrm{t}-1{)}^2}{2}+\ln \left(t+1\right)\right\}+\mathrm{c} \\ \mathrm{hence} \mathrm{f}\left(\mathrm{x}\right)=2{\left({\mathrm{x}}^{\frac{1}{4}}-1\right)}^2+4\ln \left(1+{\mathrm{x}}^{\frac{1}{4}}\right)+\mathrm{c} \\ \mathrm{f}\left(0\right)=-6\Rightarrow 2+4\mathrm{\ell n}\left(1\right)+C=-6 \\ \mathrm{C}=-8 \\ \mathrm{now} \\ \mathrm{f}\left(1\right)=4\ell n2-8 \\ =4\left(\mathrm{\ell n}2-2\right) \end{gathered}$$

$$\begin{gathered} f\left(x\right)=\int e^x\left(\frac{\left(1-x^2\right)+1}{\sqrt{1+x}·(1-x{)}^{3/2}}\right)dx \\ =\int e^x\left(\frac{\left(1-x^2\right)}{\sqrt{1+x}·(1-x{)}^{3/2}}+\frac{1}{\sqrt{1+x}·(1-x{)}^{3/2}}\right)dx \\ =\int e^x\left(\sqrt{\frac{1+x}{1-x}}+\frac{1}{\sqrt{1+x}·(1-x{)}^{3/2}}\right)dx \\ =e^x\sqrt{\frac{1+x}{1-x}}+C \\ \left[\because \int e^x\left(f\left(x\right)+f^{\text{'}}\left(x\right)\right)dx=e^{x}f\left(x\right)+C\right] \\ \because f\left(0\right)=0\Rightarrow C=-1 \\ \therefore f\left(x\right)=e^x\sqrt{\frac{1+x}{1-x}}-1 \\ \mathrm{Now}, f\left(\frac{1}{2}\right)=\sqrt{3e}-1 \end{gathered}$$

$$\begin{gathered} \int x^3\sin xdx=-x^3\cos x+\int 3x^2\cos xdx \\ =-x^3\cos x+3x^2\sin x-\int 6x\sin xdx \\ =-x^3\cos x+3x^2\sin x+6x\cos x-6\sin x+C \\ \text{ So, }g\left(x\right)=-x^3\cos x+3x^2\sin x+6x\cos x-6\sin x \\ g\left(\frac{\pi }{2}\right)=\frac{3{\pi }^2}{4}-6 \\ {g}^{\text{'}}\left(x\right)=-3x^2\cos x+x^3\sin x+3x^2\cos x+6\cos x-6\cos x \\ {g}^{\text{'}}\left(\frac{\pi }{2}\right)=\frac{{\pi }^3}{8} \\ 8\left(g\left(\frac{\pi }{2}\right)+g^{\text{'}}\left(\frac{\pi }{2}\right)\right)={\pi }^3+6{\pi }^2-48 \\ \text{ So, }\alpha +\beta -\gamma =55 \end{gathered}$$

$$\begin{gathered} \frac{2x^2+5x+9}{\sqrt{x^2+x+1}}=\frac{A(x^2+x+1)+B(2x+1)+C}{\sqrt{x^2+x+1}} \\ A = 2, B =\frac{3}{2}, C = \frac{11}{2} \\ I = \int \frac{2(x^2+x+1)+{\displaystyle \frac{3}{2}}(2x+1)+{\displaystyle \frac{11}{2}}}{\sqrt{x^2+x+1}}dx \\ I=2\int \sqrt{x^2+x+1}dx +\frac{3}{2}\int \frac{2x+1}{\sqrt{x^2+x+1}}dx+\frac{11}{2}\int \frac{1}{\sqrt{x^2+x+1}}dx \\ I = 2\int \sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}dx+\frac{3}{2}\int \frac{2x+1}{\sqrt{x^2+x+1}}dx+\frac{11}{2}\int \frac{1}{\sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}}dx \\ I = 2\left[(\frac{x}{2}+\frac{1}{4})\sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}+\frac{3}{8}\ln |x+\frac{1}{2}+\sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}|\right]+3\sqrt{x^2+x+1}+\frac{11}{2}\ln |x+\frac{1}{2}+\sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}|+C \\ I=x\sqrt{x^2+x+1}+\frac{7}{2}\sqrt{x^2+x+1}+\frac{25}{4}\ln |x+\frac{1}{2}+\sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}|+C \\ Now On comparing we get \\ \alpha =\frac{7}{2}, \beta =\frac{25}{4} \\ then \alpha +4\beta =\frac{57}{2} \end{gathered}$$

$$\begin{gathered} I\left(x\right)=\int \frac{dx}{(x-11{)}^{\frac{11}{13}}(x+15{)}^{\frac{15}{13}}}. \\ I\left(x\right)=\int \frac{dx}{\left(\frac{x-11}{x+15}{)}^{\frac{11}{13}}\right(x+15{)}^2}. \\ \text{ Let }\frac{x-11}{x+15}=t\Rightarrow \frac{26}{(x+15{)}^2}dx=dt \\ I\left(x\right)=\frac{1}{26}\int \frac{dt}{t^{11/13}}=\frac{1}{26}·\frac{t^{2/13}}{2/13}+C \\ I\left(x\right)=\frac{1}{4}{\left(\frac{x-11}{x+15}\right)}^{2/13}+C \\ \text{Now,} \\ I\left(37\right)-I\left(24\right)=\frac{1}{4}{\left(\frac{26}{52}\right)}^{2/13}-\frac{1}{4}{\left(\frac{13}{39}\right)}^{2/13} \\ =\frac{1}{4}\left(\frac{1}{2^{2/13}}-\frac{1}{3^{2/13}}\right) \\ I\left(37\right)-I\left(24\right)=\frac{1}{4}\left(\frac{1}{4^{1/13}}-\frac{1}{9^{1/13}}\right) \\ I\left(37\right)-I\left(24\right)=\frac{1}{4}\left(\frac{1}{b^{\frac{1}{13}}}-\frac{1}{c^{\frac{1}{13}}}\right) \\ \text{we get} b=4 \& c=9 \\ \text{Now,} 3\left(b+c\right)=3\left(4+9\right)=39 \end{gathered}$$

$$\begin{gathered} \int e^x\left(\frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}+\frac{{\sin }^{-1}x}{{\left(1-x^2\right)}^{3/2}}+\frac{x}{1-x^2}\right)dx \\ \text{We know} \\ \int e^x\left(f\left(x\right)+f\text{'}\left(x\right)\right)dx=e^{x}f\left(x\right)+c \\ \because \frac{d}{dx}\left(\frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}\right)=\frac{{\sin }^{-1}x}{{\left(1-x^2\right)}^{3/2}}+\frac{x}{1-x^2} \\ \Rightarrow \int e^x\left(\frac{x{\sin }^{-1}x}{\sqrt{1-x^2}}+\frac{{\sin }^{-1}x}{{\left(1-x^2\right)}^{3/2}}+\frac{x}{1-x^2}\right)dx \\ ={\mathrm{e}}^{\mathrm{x}}·\frac{{\mathrm{xsin}}^{-1}\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}}+\mathrm{c}=\mathrm{g}\left(\mathrm{x}\right)+\mathrm{C} \\ \text{on comparison} \\ \mathrm{g}\left(\mathrm{x}\right)=\frac{{\mathrm{xe}}^{\mathrm{x}}{\sin }^{-1}\mathrm{x}}{\sqrt{1-{\mathrm{x}}^2}} \\ \mathrm{g}\left(\frac{1}{2}\right)=\frac{{\mathrm{e}}^{1/2}}{2}·\frac{\frac{\mathrm{\pi }}{6}\times 2}{\sqrt{3}}=\frac{\mathrm{\pi }}{6}\sqrt{\frac{\mathrm{e}}{3}} \end{gathered}$$

$$
\begin{aligned}
& \frac{1}{3} \ln \left|\left(\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right)\right|+C \\
& \int \frac{x^2-\frac{1}{x^4} d x}{\left(\left(x^3+\frac{1}{x^3}\right)^2+1\right) \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)} \\
& \text { Put } x^3+\frac{1}{x^3}=t \\
& \left(3 x^2-\frac{3}{x^4}\right) d x=d t \\
& =\frac{1}{3} \int \frac{d t}{\left(t^2+1\right) \tan ^{-1} t} \\
& \text { Put } \tan ^{-1} t=z \\
& \frac{1}{1+t^2} d t=d z \\
& =\frac{1}{3} \int \frac{d z}{z}=\frac{1}{3} \ln \left|\left(\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right)\right|+c
\end{aligned}
$$

\(\begin{aligned}
& \int_\alpha^{\log _e 4} \frac{\mathrm{dx}}{\sqrt{\mathrm{e}^x-1}}=\frac{\pi}{6} \\
& \text { Let } \mathrm{e}^{\mathrm{x}}-1=\mathrm{t}^2 \\
& \mathrm{e}^{\mathrm{x}} \mathrm{dx}=2 \mathrm{tdt} \\
& =\int \frac{2 \mathrm{dt}}{\mathrm{t}^2+1} \\
& =2 \tan ^{-1} \mathrm{t} \\
& =\left.2 \tan ^{-1}\left(\sqrt{\mathrm{e}^{\mathrm{x}}-1}\right)\right|_\alpha ^{\log _e^4} \\
& =2\left[\tan ^{-1} \sqrt{3}-\tan ^{-1} \sqrt{\mathrm{e}^\alpha-1}\right]=\frac{\pi}{6} \\
& =\frac{\pi}{3}-\tan ^{-1} \sqrt{\mathrm{e}^\alpha-1}=\frac{\pi}{12} \\
& \Rightarrow \tan ^{-1} \sqrt{\mathrm{e}^\alpha-1}=\frac{\pi}{4} \\
& \mathrm{e}^\alpha=2 \quad \mathrm{e}^{-\alpha}=\frac{1}{2} \\
& \mathrm{x}^2-\left(2+\frac{1}{2}\right) \mathrm{x}+1=0 \\
& 2 \mathrm{x}^2-5 \mathrm{x}+2=0
\end{aligned}\)

$$\begin{gathered} f\left(x\right)=7{\tan }^{6}x(1+{\tan }^{2}x)-3{\tan }^{2}x(1+{\tan }^{2}x) \\ f\left(x\right)=se{c}^{2}x(7t{\mathrm{an}}^{6}x-3{\tan }^{2}x) \\ {I}_1=\int se{c}^{2}x(7t{\mathrm{an}}^{6}x-3{\tan }^{2}x) dx \\ let tanx=t \\ {I}_1=\frac{7t^7}{7}-3\frac{t^3}{3} \\ {I}_1=t{\mathrm{an}}^{7}x-{\tan }^{3}x+c_1 \\ {I}_2=\int xf\left(x\right)dx \\ {I}_2=\int xf\left(x\right)dx-\int 1.\int f\left(x\right)dx \\ {I}_2=x(t{\mathrm{an}}^{7}x-{\tan }^{3}x)-\int (t{\mathrm{an}}^{7}x-{\tan }^{3}x)dx \\ {I}_2=x(t{\mathrm{an}}^{7}x-{\tan }^{3}x)-\int {\tan }^{3}x({\tan }^{2}x-1)se{c}^{2}xdx \\ let \tan x=t ,se{c}^{2}xdx=dt \\ {I}_2=x(t{\mathrm{an}}^{7}x-{\tan }^{3}x)-\int t^3(t^2-1)dt \\ {I}_2=x(t{\mathrm{an}}^{7}x-{\tan }^{3}x)-\int (t^5-t^3)dt \\ {I}_2=x(t{\mathrm{an}}^{7}x-{\tan }^{3}x)-(\frac{t^6}{6}-\frac{t^4}{4}) \\ {I}_2=x(t{\mathrm{an}}^{7}x-{\tan }^{3}x)-(\frac{{\tan }^{6}x}{6}-\frac{{\tan }^{4}x}{4})+c_2 \\ now 7I_1+12I_2 \\ 7(t{\mathrm{an}}^{7}x-{\tan }^{3}x)+12\left(x\right(t{\mathrm{an}}^{7}x-{\tan }^{3}x)-(\frac{\tan x^6}{6}-\frac{\tan x^4}{4}\left)\right)+c \\ \left(12x+7\right)\left({\tan }^{7}x-{\tan }^{3}x\right)+{\tan }^{4}x\left(3-2{\tan }^{2}x\right)+c \end{gathered}$$

$$\begin{gathered} I=\int \frac{dx}{x^{1/4}\left(x^{1/4}+1\right)}, \\ \mathrm{let} {\mathrm{x}}^{1/4}=\mathrm{t} \\ I=\int \frac{4t^{3}dt}{t(t+1)}=4\int \frac{t^2}{t+1}dt \\ =4\int (t-1)dt+4\int \frac{1}{t+1}dt \\ \Rightarrow 4\left(\frac{t^2}{2}-t\right)+4\ln \left|t+1\right|+c \\ f\left(x\right)=2\left(x^{\frac{1}{2}}\right)-4x^{\frac{1}{4}}+4\ln \left|1+x^{\frac{1}{4}}\right|+c \\ f\left(0\right)=-6 \Rightarrow c=-6 \\ f\left(2\right)=2\left(\sqrt{2}\right)-4.2^{\frac{1}{4}}+4\ln \left|1+2^{\frac{1}{4}}\right|-6 \\ \Rightarrow f\left(2\right)=4\left[\frac{1}{\sqrt{2}}-2^{1/4}+\ln \left|1+2^{1/4}\right|\right]-6 \end{gathered}$$

$$\begin{gathered} \frac{2x^2+5x+9}{\sqrt{x^2+x+1}}=\frac{A(x^2+x+1)+B(2x+1)+C}{\sqrt{x^2+x+1}} \\ A = 2, B =\frac{3}{2}, C = \frac{11}{2} \\ I = \int \frac{2(x^2+x+1)+{\displaystyle \frac{3}{2}}(2x+1)+{\displaystyle \frac{11}{2}}}{\sqrt{x^2+x+1}}dx \\ I=2\int \sqrt{x^2+x+1}dx +\frac{3}{2}\int \frac{2x+1}{\sqrt{x^2+x+1}}dx+\frac{11}{2}\int \frac{1}{\sqrt{x^2+x+1}}dx \\ I = 2\int \sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}dx+\frac{3}{2}\int \frac{2x+1}{\sqrt{x^2+x+1}}dx+\frac{11}{2}\int \frac{1}{\sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}}dx \\ I = 2\left[(\frac{x}{2}+\frac{1}{4})\sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}+\frac{3}{8}\ln |x+\frac{1}{2}+\sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}|\right]+3\sqrt{x^2+x+1}+\frac{11}{2}\ln |x+\frac{1}{2}+\sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}|+C \\ I=x\sqrt{x^2+x+1}+\frac{7}{2}\sqrt{x^2+x+1}+\frac{25}{4}\ln |x+\frac{1}{2}+\sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}|+C \\ Now On comparing we get \\ \alpha =\frac{7}{2}, \beta =\frac{25}{4} \\ then \alpha +4\beta =\frac{57}{2} \end{gathered}$$