Question

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

Answer 1

Let $I=\int \frac{2x^2+1}{x^2(x^2+4)}dx$
Perform partial fraction decomposition:
$\int \frac{2x^2+1}{x^2(x^2+4)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+4}$

$\frac{2x^2+1}{x^2(x^2+4)}=\frac{x^2(Cx+D)+x(x^2+4)A+(x^2+4)B}{x^2(x^2+4)}$

After simplifying, we get
Coefficients near like terms should be equal, so the following system is obtained:
$A+C=0,B+D=2,4A=0,4B=1$.

Solve it, then A = 0, $B=\frac{1}{4}$, C = 0, $D=\frac{7}{4}$
$2x^2+1=x^3(A+C)+x^2(B+D)+4xA+4B$

Therefore,

$\frac{2x^2+1}{x^2(x^2+4)}=\frac{1}{4}\frac{1}{x^2}+\frac{7}{4(x^2+4)}$

Integrating separately, we get
$=\frac{1}{4}\int \frac{1}{x^2}dx+\frac{7}{4}\int \frac{1}{x^2+4}$

We know that, $\int \frac{1}{x^2+1}={\tan }^{-1}x$
$=\frac{1}{4}\left(\frac{x^{-2+1}}{-2+1}\right)+\frac{7}{4\times 2}\left({\tan }^{-1}\left(\frac{x}{2}\right)\right)+C$

$=\frac{1}{4}\left(\frac{1}{-x}\right)+\frac{7}{8}\left({\tan }^{-1}\left(\frac{x}{2}\right)\right)+C$

$=\frac{1}{x}[-\frac{1}{4}+\frac{7}{8}x{\tan }^{-1}\left(\frac{x}{2}\right)]+C$