Find $\int \frac {\mathrm dx}{(x + 1)(x^2 + 2)}$ I'm supposed to find the antiderivative of

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

and I'm completely stumped. I've been trying substitutions with $u = x^2$ and that's led me nowhere. I don't think I can use partial fractions here since I have one linear factor and one quadratic factor below the division line, right?

Consider $$\int {1\over (x+1)(x^2+2)}dx.$$
Using the method of partial fraction decomposition we have $${1\over (x+1)(x^2+2)} = {A\over x+1}+{Bx+C\over x^2+2}.$$ Solving for constans $A$, $B$,and $C$ we obtain $${1\over 3(x+1)}+{1-x\over 3(x^2+2)}.$$ So we must integrate $$\int{1\over 3(x+1)}+{1-x\over 3(x^2+2)}dx.$$ Rewrite the integral as $$\int{1\over 3}\cdot {1\over x+1}+{1\over 3}\cdot {1\over x^2+2}-{1\over 3}\cdot {x\over x^2+2}dx.$$ The first term is a $u$-substitution, the second term involves a $\tan^{-1}(x)$ identity and the third term is another $u$-substitution. Thus we integrate and obtain $${1\over 3}\cdot \ln(x+1)+{1\over 3\sqrt2}\cdot \tan^{-1}({x\over \sqrt2})-{1\over 6}\cdot \ln(x^2+2)+C$$ where $C$ is a constant.
