# The integral

Question:

The integral $\int \frac{3 x^{13}+2 x^{11}}{\left(2 x^{4}+3 x^{2}+1\right)^{4}} d x$ is equal to:

(where $\mathrm{C}$ is a constant of integration)

1. (1) $\frac{x^{4}}{6\left(2 x^{4}+3 x^{2}+1\right)^{3}}+\mathrm{C}$

2. (2) $\frac{x^{12}}{6\left(2 x^{4}+3 x^{2}+1\right)^{3}}+\mathrm{C}$

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

4. (4) $\frac{x^{12}}{\left(2 x^{4}+3 x^{2}+1\right)^{3}}+C$

Correct Option: , 2

Solution:

$I=\int \frac{3 x^{13}+2 x^{11}}{\left(2 x^{4}+3 x^{2}+1\right)^{4}} d x=\int \frac{3 x^{13}+2 x^{11}}{x^{16}\left(2+\frac{3}{x^{2}}+\frac{1}{x^{4}}\right)^{4}} d x$

$I=\int \frac{\frac{3}{x^{3}}+\frac{2}{x^{5}}}{\left(2+\frac{3}{x^{2}}+\frac{1}{x^{4}}\right)^{4}} d x$

Let $\quad 2+\frac{3}{x^{2}}+\frac{1}{x^{4}}=\mathrm{t},-2\left(\frac{3}{x^{3}}+\frac{2}{x^{5}}\right) d x=d t$

Then, $\quad I=\int \frac{-\frac{d t}{2}}{t^{4}}=-\frac{1}{2} \frac{t^{-4+1}}{-4+1}+C$

$I=\frac{-1}{2} \times \frac{1}{(-3)} \frac{1}{\left(2+\frac{3}{x^{2}}+\frac{1}{x^{4}}\right)^{3}}+C$

$I=\frac{1}{6} \frac{x^{12}}{\left(2 x^{4}+3 x^{2}+1\right)^{3}}+C$