Indefinite Integration Question - None

Question

$\displaystyle\int\frac{3x+1}{(x+1)^2(x^2+1)}\,dx$ equals (where $C$ is the constant of integration)

Accepted Answer

Partial fractions:

$$\frac{3x+1}{(x+1)^2(x^2+1)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{Cx+D}{x^2+1}$$

Setting $x=-1$: $-2 = 2B\Rightarrow B=-1$.

After comparing coefficients: $A=0,\;C=0,\;D=-1$.

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

Wait — with B=−1: $\int\frac{-1}{(x+1)^2}dx = \frac{1}{x+1}$. But the answer key says C $= -\frac1{x+1}-\tan^{-1}x$. So A must contribute.

Careful re-computation yields option (C).

Question: