The value of the integral \({\int }_{-1}^{2}{\log }_{e}\left(x+\sqrt{{x}^{2}+1}\right)dx\) is [JEE Main 2024, 9 Apr (Shi…

Let \(I=\int_{-1}^{2}\log_e\left(x+\sqrt{x^2+1}\right)dx\).

Now \( \frac{d}{dx}\log_e\left(x+\sqrt{x^2+1}\right)=\frac{1}{\sqrt{x^2+1}} \).

So \(I=\left[x\log_e\left(x+\sqrt{x^2+1}\right)\right]_{-1}^{2}-\int_{-1}^{2}\frac{x}{\sqrt{x^2+1}}dx\).

Since \( \int \frac{x}{\sqrt{x^2+1}}dx=\sqrt{x^2+1} \), we get

\(I=\left[x\log_e\left(x+\sqrt{x^2+1}\right)-\sqrt{x^2+1}\right]_{-1}^{2}\).

At \(x=2\), the value is \(2\log_e(2+\sqrt{5})-\sqrt{5}\).

At \(x=-1\), the value is \(-\log_e(\sqrt{2}-1)-\sqrt{2}\).

Therefore \(I=2\log_e(2+\sqrt{5})-\sqrt{5}+\log_e(\sqrt{2}-1)+\sqrt{2}\).

Now \(2\log_e(2+\sqrt{5})=\log_e(2+\sqrt{5})^2=\log_e(9+4\sqrt{5})\).

Also \( \sqrt{2}-1=\frac{1}{\sqrt{2}+1} \), so \( \log_e(\sqrt{2}-1)=-\log_e(\sqrt{2}+1) \).

Hence \(I=\sqrt{2}-\sqrt{5}+\log_e(9+4\sqrt{5})-\log_e(1+\sqrt{2})\).

So \(I=\sqrt{2}-\sqrt{5}+\log_e\left(\frac{9+4\sqrt{5}}{1+\sqrt{2}}\right)\).