Question

$\int \frac{x^5}{x^2+9}dx$

Answer 1

Consider the given integral.

$I=\int \frac{x^5}{x^2+9}dx$

Let,

$t=x^2+9$

$dt=2xdx$

Therefore,

$I=\frac{1}{2}\int \frac{{(t-9)}^2}{t}dt$

$I=\frac{1}{2}\int \frac{t^2+81-18t}{t}dt$

$I=\frac{1}{2}\int (t+\frac{81}{t}-18)dt$

$I=\frac{1}{2}[\frac{t^2}{2}+81\ln \left(t\right)-18t]+C$

Put the value of $t$, we get

$I=\frac{1}{2}[\frac{{(x^2+9)}^2}{2}+81\ln (x^2+9)-18(x^2+9)]+C$

Hence, the value of this expression is $\frac{1}{2}[\frac{{(x^2+9)}^2}{2}+81\ln (x^2+9)-18(x^2+9)]+C$