Q114 / 160 · TS EAMCET 2024 (11 May Shift 1) · Mathematics
$\int \frac{d x}{(x+1) \sqrt{x^2+4}}=$
- A$\frac{1}{2} \sqrt{\frac{x+1}{x+2}}+c$
- B$\log \left|\frac{x+2}{x+1}\right|+c$
- Correct answerC$-\frac{1}{\sqrt{5}} \sinh ^{-1}\left(\frac{4-x}{2(x+1)}\right)+c$
- D$-\frac{1}{\sqrt{5}} \cosh ^{-1}\left(\frac{4+x}{2(x-1)}\right)+c$
Solution
$\begin{aligned} & \text { Let } K=\int \frac{d x}{(x+1) \sqrt{x^2+4}} \\ & \text { Let } x+1=\frac{1}{t} \Rightarrow x=\frac{1}{t}-1=\frac{1-t}{t} \Rightarrow d x=-\frac{1}{t^2} d t \\ & I=-\int \frac{\frac{1}{t^2} d t}{\frac{1}{t} \sqrt{\left(\frac{1-t}{t}\right)^2+4}}=-\int \frac{\sqrt{t^2} d t}{t \sqrt{(1-t)^2+4 t^2}} \\ & =-\int \frac{d t}{\sqrt{\left(1-t^2\right)+4 t^2}}=-\int \frac{d t}{\sqrt{5 t^2-2 t+1}} \times \frac{\frac{\sqrt{5}}{\frac{\sqrt{5}}{2}}}{2} \\ & =\frac{-\sqrt{5}}{2} \int \frac{d t}{\sqrt{\frac{25}{4} t^2-\frac{5}{2} t+\frac{5}{4}}}=\frac{-\sqrt{5}}{2} \int \frac{d t}{\sqrt{\frac{25}{4}} t^2+\frac{1}{4}-\frac{5}{2} t+1} \\ & =\frac{-\sqrt{5}}{2} \int \frac{d t}{\sqrt{\left(\frac{5}{2} t-\frac{1}{2}\right)^2+1}} \\ & =\frac{-\sqrt{5}}{2} \times \frac{2}{5} \sinh ^{-1}\left(\frac{5 t}{2}-\frac{1}{2}\right)+C\left[\because \int \frac{d t}{\sqrt{x^2+1}}=\sinh ^{-1} x\right] \\ & I=\frac{-1}{\sqrt{5}} \sinh ^{-1}\left(\frac{5 t-1}{2}\right)+C \text { Put } t=\frac{1}{x+1}\end{aligned}$
$I=\frac{-1}{\sqrt{5}} \sinh ^{-1}\left(\frac{\frac{5}{x+1}-1}{2}\right)+C=\frac{-1}{\sqrt{5}} \sinh ^{-1}\left(\frac{4-x}{2(x+1)}\right)+C$.
Hence, the correct answer is option: 'C'.