Question

$\int \frac{x^{9}dx}{(4x^2+1{)}^6}$ is equal to
(where ${}^{\prime }{C}^{\prime }$ is the constant of integration)

• A. $\frac{x^{10}}{50}\cdot \frac{1}{(4x^2+1{)}^5}+C$
• B. $-\frac{x^{10}}{10}\cdot \frac{1}{(4x^2+1{)}^5}+C$
• C. $\frac{1}{10x^2}\cdot \frac{1}{{(4x^2+1)}^5}+C$
• D. $\frac{x^{10}}{10}\cdot \frac{1}{(4x^2+1{)}^5}+C$

Answer 1

The correct option is D $\frac{x^{10}}{10}\cdot \frac{1}{(4x^2+1{)}^5}+C$

$I=\int \frac{x^{9}dx}{(4x^2+1{)}^6}$
$=\int \frac{x^{9}dx}{x^{12}{(4+\frac{1}{x^2})}^6}$
$=\int \frac{dx}{x^3{(4+\frac{1}{x^2})}^6}$
$=-\frac{1}{2}\int \frac{d(4+\frac{1}{x^2})}{{(4+\frac{1}{x^2})}^6}$
$=-\frac{1}{2}\frac{{(4+\frac{1}{x^2})}^{-5}}{-5}+C$
$=\frac{1}{10}{(4+\frac{1}{x^2})}^{-5}+C$