If \(f(x)=\int \frac{1}{x^{1 / 4}\left(1+x^{1 / 4}\right)} \mathrm{d} x, f(0)=-6\), then \(f(1)\) is equal to : [JEE Mai…

\(\text{let}x={t}^{4}\\ dx=4{t}^{3}dt\\ \text{then}\int \frac{1}{{x}^{\frac{1}{4}}\left(1+{x}^{\frac{1}{4}}\right)}dx\\ =\int \frac{4{t}^{3}dt}{t(1+t)}\\ =\int \frac{4{\mathrm{t}}^{2}}{1+\mathrm{t}}\mathrm{dt}\\ =4\int \frac{\left({\mathrm{t}}^{2}-1\right)+1}{1+\mathrm{t}}\mathrm{dt}\\ =4\int \left(\mathrm{t}-1\right)+\frac{1}{\mathrm{t}+1}\mathrm{dt}\\ =4\left{\frac{(\mathrm{t}-1{)}^{2}}{2}+\ln \left(t+1\right)\right}+\mathrm{c}\\ \mathrm{hence}\mathrm{f}\left(\mathrm{x}\right)=2{\left({\mathrm{x}}^{\frac{1}{4}}-1\right)}^{2}+4\ln \left(1+{\mathrm{x}}^{\frac{1}{4}}\right)+\mathrm{c}\\ \mathrm{f}\left(0\right)=-6\Rightarrow 2+4\mathrm{ℓn}\left(1\right)+C=-6\\ \mathrm{C}=-8\\ \mathrm{now}\\ \mathrm{f}\left(1\right)=4ℓn2-8\\ =4\left(\mathrm{ℓn}2-2\right)\)