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

Question

Question : Evaluate $$\int \frac {2x}{(x^2+1)(x^2+2)^2} dx $$

I am not able to find the value of the following integrals I tried with the method of substituting the denominator by $t$ but I cannot find the answer. Any help/hint would be appreciated.

Answer 1

Substituting $t=x^2$ the integrand takes the form $$\displaylines{{1\over (t+1)(t+2)^2} =\left [{1\over t+1}- {1\over t+2}\right ] {1\over t+2}\\ = \left[ {1\over t+1}- {1\over t+2}\right ]-{1\over (t+2)^2}}$$ Thus the result is equal $$\log {x^2+1\over x^2+2}+{1\over x^2+2}$$

Answer 2

HINT

Let $u = x^{2} + 1$. Then the proposed indefinite integral reduces to \begin{align*} \int\frac{2x}{(x^{2} + 1)(x^{2} + 2)^{2}}\mathrm{d}x & = \int\frac{\mathrm{d}(x^{2} + 1)}{(x^{2} + 1)((x^{2} + 1) + 1)^{2}}= \int\frac{\mathrm{d}u}{u(u+1)^{2}} \end{align*}

Can you take it from here?

Answer 3

Integrate by parts as follows

\begin{align} &\int \frac {2x}{(x^2+1)(x^2+2)^2} dx = \int \frac1{x^2+1}d\left( \frac{x^2+1}{x^2+2}\right)\\ =& \ \frac1{x^2+2}+\int \frac{2x}{x^2+1} -\frac{2x}{x^2+2}\ dx = \frac1{x^2+2} +\ln \frac{x^2+1}{x^2+2}+C \end{align}