Question

Find $\int \frac{2x}{(x^2+1)(x^4+4)}dx.$

Answer 1

We have,

$\int \frac{2x}{(x^2+1)(x^4+4)}dx$

$\Rightarrow \int \frac{2x}{(x^2+1)({\left(x^2\right)}^2+4)}dx$

Let

$x^2=t$

$2xdx=dt$

Using by partial dfraction and we get,

$\frac{1}{(t+1)(t^2+4)}=\frac{A}{(t+1)}+\frac{Bt+C}{(t^2+4)}......\left(A\right)$

$1=A(t^2+4)+(Bt+C)(t+1)$

$1=A{t}^2+4A+B{t}^2+Ct+Bt+C$

$1=(A+B)t^2+(B+C)t+(4A+C)$

Comparing coefficients and we get,

$A+B=0.......\left(1\right)$

$B+C=0.......\left(2\right)$

$4A+C=1.......\left(3\right)$

By equation (1), (2) and (3) to,

$A=\frac{1}{5},B=-\frac{1}{5},C=\frac{1}{5}$

Then, by equation (A) and we get,

$\frac{1}{(t+1)(t^2+4)}=\frac{1}{5(t+1)}+\frac{-\frac{1}{5}t+\frac{1}{5}}{(t^2+4)}$

$\frac{1}{(t+1)(t^2+4)}=\frac{1}{5(t+1)}-\frac{1}{5}\frac{t}{(t^2+4)}+\frac{1}{5(t^2+4)}$

On integrating and we get,

$\int \frac{1}{(t+1)(t^2+4)}=\frac{1}{5}\int \frac{1}{(t+1)}dt-\frac{1}{5}\int \frac{t}{(t^2+4)}dt+\frac{1}{5}\int \frac{1}{(t^2+2^2)}dt$

$=\frac{1}{5}\log t-\frac{1}{5}\log (t^2+4)+\frac{1}{5}\times \frac{1}{2}{\tan }^{-1}\frac{t}{2}+C$

$=\frac{1}{5}\log \frac{t}{t^2+4}+\frac{1}{10}{\tan }^{-1}\frac{t}{2}+C$

Put the value of $t=x^2$ and we get,

$=\frac{1}{5}\log \frac{x^2}{x^4+4}+\frac{1}{10}{\tan }^{-1}\frac{x^2}{2}+C$

Hence, this is the answer.