Question

The value of $\int \frac{3x^{13}+2x^{11}}{(2x^4+3x^2+1{)}^4}dx$ is equal to
(where $C$ is constant of integration )

• A. $\frac{x^4}{6(2x^4+3x^2+1{)}^3}+C$
• B. $\frac{x^{12}}{6(2x^4+3x^2+1{)}^3}+C$
• C. $\frac{x^4}{(2x^4+3x^2+1{)}^3}+C$
• D. $\frac{x^{12}}{(2x^4+3x^2+1{)}^3}+C$

Answer 1

The correct option is B $\frac{x^{12}}{6(2x^4+3x^2+1{)}^3}+C$

Let $I=\int \frac{3x^{13}+2x^{11}}{(2x^4+3x^2+1{)}^4}dx$
$\Rightarrow I=\int \frac{3x^{13}+2x^{11}}{x^{16}{(2+\frac{3}{x^2}+\frac{1}{x^4})}^4}dx$
$\Rightarrow I=\int \frac{\frac{3}{x^3}+\frac{2}{x^5}}{{(2+\frac{3}{x^2}+\frac{1}{x^4})}^4}dx$
Put $2+\frac{3}{x^2}+\frac{1}{x^4}=t$
$\Rightarrow -2(\frac{3}{x^3}+\frac{2}{x^5})dx=dt$
$\therefore I=-\frac{1}{2}\int \frac{dt}{t^4}$
$\Rightarrow I=\frac{1}{6t^3}+C$
$\Rightarrow I=\frac{1}{6{(2+\frac{3}{x^2}+\frac{1}{x^4})}^3}+C$
$\therefore I=\frac{x^{12}}{6(2x^4+3x^2+1{)}^3}+C$