Mathematics

Evaluate: $$\displaystyle \int \dfrac{1}{(x + 1)\sqrt{x^2 - 1}}dx$$


ANSWER

$$\dfrac {\sqrt {x-1}}{\sqrt{x+1}}+c$$


SOLUTION
$$\displaystyle\int \dfrac{1}{(x-1)\sqrt{x^2-1}}dx$$
Substitute $$x=\sec u$$
$$dx=\sec u+\tan u du$$
$$=\dfrac{\sec u \tan u}{(\sec u+1)\sqrt{\sec^2y-1}}du$$
$$\sec^2u-1=\tan^2u$$
$$=\dfrac{\sec u}{\sec u+1}du=\displaystyle\int \left(\dfrac{\sec u+1}{\sec u+1}-\dfrac{1}{\sec u+1}\right)du$$
$$=\displaystyle\int \left(1-\dfrac{1}{\sec u +1}\right)du$$
$$=u-\displaystyle\int \dfrac{1}{\dfrac{\tan^2\dfrac{u}{2}+1}{1-\tan^2\dfrac{u}{2}}+1}du$$
$$v=\tan \dfrac{u}{2}$$
$$\Rightarrow du=\dfrac{2}{\sec^2\dfrac{u}{2}}$$
$$du=\dfrac{2}{v^2+1}dv$$
$$=u-\displaystyle\int \dfrac{(v-1)(v+1)}{v^2+1}dv$$
$$=u-\displaystyle\int \dfrac{v^2+1}{v^2+1}-\dfrac{2}{v^2+1}dv=u-\displaystyle\int 1-\dfrac{2}{v^2+1}dv$$
$$=u-(v-2\tan^{-1}(v))$$
Resubstituting $$v=\tan \dfrac{u}{2}$$
$$=u-\left(\tan \dfrac{u}{2}-2\tan^{-1}\left(\tan\dfrac{u}{2}\right)\right)$$
$$=\tan\dfrac{u}{2}$$
Resubstituting $$u=\sec^{1}(x)$$
$$\tan \left(\dfrac{\sec^{-1}(x)}{2}\right)=\dfrac{\sqrt{x-1}}{\sqrt{x+1}}$$
$$\therefore \displaystyle\int \dfrac{1}{(x+1)\sqrt{x^2-1}}dx=\dfrac{\sqrt{x-1}}{\sqrt{x+1}}+C$$.
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