Evaluate the following integral:

Question:

Evaluate the following integral:

$\int \frac{x^{2}}{\left(x^{2}+1\right)\left(3 x^{2}+4\right)} d x$

Solution:

$I=\int \frac{x^{2}}{\left(x^{2}+1\right)\left(3 x^{2}+4\right)} d x$

$\frac{x^{2}}{\left(x^{2}+1\right)\left(3 x^{2}+4\right)}=\frac{A x+B}{x^{2}+1}+\frac{C x+D}{3 x^{2}+4}$

$x^{2}=(A x+B)\left(3 x^{2}+4\right)+(C x+D)\left(x^{2}+1\right)$

$=(3 A+C) x^{3}+(3 B+D) x^{2}+(4 A+C) x+4 B+D$

Equating similar terms

$3 A+C=0$

$3 B+D=1$

$4 A+C=0$

$4 B+D=0$

Solving we get,

$A=0, B=-1, C=0, D=4$

Thus,

$I=\int \frac{-d x}{x^{2}+1}-\int \frac{4 d x}{3 x^{2}+4}$

$I=-\tan ^{-1} x+\frac{4}{3} \int \frac{d x}{x^{2}+\left(\frac{2}{\sqrt{3}}\right)^{2}}$

$I=-\tan ^{-1} x+\frac{4}{3} \cdot \frac{\sqrt{3}}{2} \tan ^{-1} \frac{\sqrt{3} x}{2}+C$

$I=\frac{2}{\sqrt{3}} \tan ^{-1} \frac{\sqrt{3} x}{2}-\tan ^{-1} x+C$

Leave a comment

None
Free Study Material