Question

If $I=\int \frac{5x^8+7x^6}{{(x^2+1+2x^7)}^2}dx$ then $I$ is equal to

• A. $\frac{x^2}{2x^2+x^2+1}+C$
• B. $\frac{x^5}{x^2+1+2x^7}+C$
• C. $\frac{-1}{2x^7+x^2+1}+C$
• D. $\frac{p\left(X\right)}{q\left(x\right)},deg$ p(x)$=deg$ q(x)$=7$

Answer 1

The correct option is D $\frac{p\left(X\right)}{q\left(x\right)},deg$ p(x)$=deg$ q(x)$=7$

$I=\int \frac{5x^8+7x^6}{{(x^2+1+2x^7)}^2}dx=\int \frac{5x^8+7x^6}{x^{14}{\left(\frac{1}{x^5}+\frac{1}{x^7}+2\right)}^2}dx$
$=\int \frac{\frac{5}{x^6}+\frac{7}{x^8}}{{(\frac{1}{x^5}+\frac{1}{x^7}+2)}^2}dx$
Substitute $t=(\frac{1}{x^5}+\frac{1}{x^7}+2)\Rightarrow dt=(-\frac{5}{x^6}-\frac{7}{x^8})dx$
$I=-\int \frac{dt}{t^2}=\frac{1}{t}+c=\frac{x^7}{x^2+1+2x^7}+c$