$\int \sqrt{x-1}(x \sqrt{x+1})^{-1} d x=$

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Aicharya · Accepted Answer

$I=\int \sqrt{x-1}(x \sqrt{x+1})^{-1} d x=\int \frac{1}{x} \cdot \sqrt{\frac{x-1}{x+1}} d x$ Put $\frac{x-1}{x+1}=t^2 \Rightarrow x=\frac{1+t^2}{1-t^2}$ So, $d x=\frac{\left(1-t^2\right)(2 t)-\left(l+t^2\right)(-2 t)}{\left(1-t^2\right)^2} d t$ $\Rightarrow d x=\frac{4 t}{\left(1-t^2\right)^2} d t$ So, $\quad I=\int\left(\frac{1-t^2}{1+t^2}\right) t \frac{4 t}{\left(1-t^2\right)^2} d t=\int \frac{4 t^2}{\left(1+t^2\right)\left(1-t^2\right)} d t$ $\begin{aligned} & =2 \int\left(\frac{1}{1-t^2}-\frac{1}{\left.1+t^2\right)}\right) d t \ & =2\left[-\frac{1}{2} \log \left(\frac{1-t}{1+t}\right)-\tan ^{-1}(t)\right]+c \end{aligned}$ $=-\log _e\left(\frac{1-\sqrt{\frac{x-1}{x+1}}}{1+\sqrt{\frac{x-1}{x+1}}}\right)-2 \tan ^{-1} \sqrt{\frac{x-1}{x+1}}+c$ $=-\log _e\left(\frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}\right)-\tan ^{-1}\left(\frac{2 \sqrt{\frac{x-1}{x+1}}}{1-\frac{x-1}{x+1}}\right)+c$ $=-\log _e\left(\frac{x+1+x-1-2 \sqrt{x^2-1}}{(x+1)-(x-1)}\right)-\tan ^{-1} \sqrt{x^2-1}+c$ $=-\log _e\left(x-\sqrt{x^2-1}\right)-\sec ^{-1} x+c$ $=\log _e\left(x+\sqrt{x^2-1}\right)-\sec ^{-1} x+c$

\(\int \sqrt{x-1}(x \sqrt{x+1})^{-1} d x=\)

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