JEE MainMathematicsToughIndefinite integration

$\int\frac{x^{2} - 1}{\left( x^{2} + 1 \right)\sqrt{1 + x^{4}}}$
dx is equal to

Show answer and solution

I = \int\frac{x^{2} - 1}{\left( x^{2} + 1 \right)\sqrt{x^{4} + 1}}dx = \int\frac{x^{2}\left( 1 - \frac{1}{x^{2}} \right)}{x^{2}\left( x + \frac{1}{x} \right)\sqrt{x^{2} + \frac{1}{x^{2}}}}dx

= \int\frac{\left( 1 - \frac{1}{x^{2}} \right)dx}{\left( x + \frac{1}{x} \right)\sqrt{(x + 1/x)^{2} - 2}}

Putting

\ x + \frac{1}{x} = t,

we
have

I = \int\frac{dt}{t\sqrt{t^{2} - 2}}

\

Again putting

t^{2} - 2 = y^{2},2tdt = 2ydy

I = \int\frac{ydy}{\left( y^{2} + 2 \right)y} = \frac{1}{\sqrt{2}}\tan^{- 1}\frac{y}{\sqrt{2}} = \frac{1}{\sqrt{2}}\tan^{- 1}\frac{\sqrt{x^{2} + 1/x^{2}}}{\sqrt{2}} + c

Source: Grass-G / M-12 Mathematics Paper 2 Nov.docx

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.