Find $\int\frac{x^2+1}{(x^2+2)(x^2+3)}dx$.

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

The correct answer is Option (2) → $\frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{x}{\sqrt{3}}\right) - \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x}{\sqrt{2}}\right) + C$

$I=\int\frac{x^2+1}{(x^2+2)(x^2+3)}dx$

Putting $x^2 = y$

$\frac{x^2 + 1}{(x^2 + 2)(x^2 + 3)} = \frac{y + 1}{(y + 2)(y + 3)}$

We can write this in form

$\frac{y + 1}{(y + 2)(y + 3)} = \frac{A}{(y + 2)} + \frac{B}{(y + 3)}$

$\frac{y + 1}{(y + 2)(y + 3)} = \frac{A(y + 3) + B(y + 2)}{(y + 2)(y + 3)}$

$y + 1 = A(y + 3) + B(y + 2)$

Putting $y = -3$

$-3 + 1 = A(-3 + 3) + B(-3 + 2)$

$-2 = A \times 0 + B \times -1$

$-2 = -B$

$B = 2$

Putting $y = -2$

$-2 + 1 = A(-2 + 3) + B(-2 + 2)$

$-1 = A \times 1 + B \times 0$

$-1 = A$

$A = -1$

Hence we can write

$\frac{y + 1}{(y + 2)(y + 3)} = \frac{-1}{(y + 2)} + \frac{2}{(y + 3)}$

Substituting back $y = x^2$

$\frac{x^2 + 1}{(x^2 + 2)(x^2 + 3)} = \frac{-1}{(x^2 + 2)} + \frac{2}{(x^2 + 3)}$

Therefore,

$\int \frac{x^2 + 1}{(x^2 + 2)(x^2 + 3)} \, dx = \int \frac{-1}{(x^2 + 2)} \, dx + \int \frac{2}{(x^2 + 3)} \, dx$

$= -\int \frac{1}{x^2 + (\sqrt{2})^2} \, dx + 2 \int \frac{1}{x^2 + (\sqrt{3})^2} \, dx$

By using formula

$\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C$

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