Question

Integrate the rational function:

${\displaystyle \frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}}$

Answer 1

${\displaystyle \frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}}$  Taking x² = y

${\displaystyle \frac{(y+1)(y+2)}{(y+3)(y+4)}=\frac{y^2+3y+2}{y^2+7y+12}}$

${\displaystyle =1-\frac{4y+10}{y^2+7y+12}}$

${\displaystyle =1-\frac{4y+10}{(y+3)(y+4)}}$

Let ${\displaystyle \frac{4y+10}{(y+3)(y+4)}=\frac{A}{(y+3)}+\frac{B}{y+4}}$

4y + 10 = A (y + 4) + B (y + 3)

Putting y = -4 - 6 = 0 - B

⇒ B = 6

Putting y = -3, -2 = A + 0

⇒  A = -2

${\displaystyle \therefore \frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}=1-[\frac{-2}{y+3}+\frac{6}{y+4}]}$

${\displaystyle =1+\frac{2}{y+3}+\frac{6}{y+4}}$

${\displaystyle \int \frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}}$ dx

${\displaystyle =\int dx+2\int \frac{1}{x^2\sqrt{3^2}}+6\int \frac{1}{x^2+4}}$ dx

${\displaystyle =x+\frac{2}{\sqrt{3}} \frac{{\tan }^{-1}x}{\sqrt{3}}-\frac{6}{2} {\tan }^{-1}\left(\frac{x}{2}\right)+C}$

${\displaystyle =x+\frac{2}{\sqrt{3}} {\tan }^{-1} \frac{x}{\sqrt{3}}-3 {\tan }^{-1} \frac{x}{2}+C}$