How do I integrate $\frac{1}{(x+1)\sqrt{x^2-1}}$ and $\frac{1}{x^2\sqrt{x^2+1}}$?

Question

How do I integrate

1. $\Large\frac{1}{(x+1)\sqrt{x^2-1}}$?

2. $\Large\frac{1}{x^2\sqrt{x^2+1}}$

How do I approach?

I'm give an answer

1. $\Large\sqrt{\frac{x-1}{x+1}}$+C

2. $\Large\frac{\sqrt{x^2+1}}{x}$+C

Answer 1

Let's try to bring some familiar forms out of the original function: $$\frac{1}{(x+1)\sqrt{x^2-1}}=\frac{1}{(x+1)\sqrt{x-1}\sqrt{x+1}}=\frac{1}{(x+1)^2\sqrt{\dfrac{{x-1}}{{x+1}}}}$$ $$u=\sqrt{\frac{x-1}{x+1}}\implies du={\dfrac{x+1-(x-1)}{(x+1)^2}\over 2\sqrt{\dfrac{x-1}{x+1}}}dx=\frac{1}{(x+1)^2\sqrt{\dfrac{{x-1}}{{x+1}}}}dx$$

So the first integral is $$\int\frac{1}{(x+1)^2\sqrt{\dfrac{{x-1}}{{x+1}}}}dx=\int du=u+C=\boxed{\color{blue}{\sqrt{\frac{x-1}{x+1}}+C}}$$

As for the second: $$u=\frac 1x\implies du=-\frac 1{x^2}dx$$ $$\int\frac{1}{x^2\sqrt{x^2+1}}dx=-\int\frac{udu}{\sqrt{1+u^2}}$$

Then set $$y=1+u^2\implies dy=2udu$$ $$-\int\frac 12\frac 1{\sqrt y}dy=\sqrt{y}+C=-\sqrt{1+u^2}+C=-\sqrt{1+\frac{1}{x^2}}+C=\boxed{\color{blue}{-\frac{\sqrt{x^2+1}}x +C}}$$ (Notice that in the question, the minus sign is missing from the proposed answer for the second integral)

Answer 2

Using Trigonometric substitutions,

for the first case set $x=\sec y$ where $0\le y<\dfrac\pi2$ $\implies dx=\sec y\tan y\ dy$

$$\int\frac{dx}{(x+1)\sqrt{x^2-1}}=\int\dfrac{\sec y\tan y\ dy}{(\sec y+1)\tan y}=\int\dfrac{dy}{1+\cos y}=\dfrac12\int\sec^2\dfrac y2\ dy=\tan\dfrac y2+K$$

Now use $\tan\dfrac y2=\dfrac{\sin y}{1+\cos y}$

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For the second, set $x=\tan u$ where $-\dfrac\pi2<u<\dfrac\pi2$

$$\int\dfrac{dx}{x^2\sqrt{x^2+1}}=\int\dfrac{\sec^2u\ du}{\sec u\tan^2u}=\int\dfrac{\cos u\ du}{\sin^2u}$$

Set $\sin u=v$

So, we can start with $v=\sin u=\dfrac x{\sqrt{x^2+1}}$