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

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

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

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

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

$= \tan^{-1} x + \int \frac{-\frac{dt}{2}}{t}$

$\left(\int\frac{dx}{1+x^2}=\tan^{-1}x\, \text{and by taking} 1 + \frac{1}{x^2} = t \Rightarrow \frac{-2}{x^3} dx = dt \right)$

$= \tan^{-1} x - \frac{1}{2} \ln \left( 1 + \frac{1}{x^2} \right) + C$   $\left(∴t=1+ \frac{1}{x^2}\right)$

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