Question
Evaluate $\int_{-1}^1 5 x^4 \sqrt{x^5+1} d x$.

Answer

Let
$I=\int_{-1}^1 5 x^4 \sqrt{x^5+1} d x$
Putting $t=x^5+1 \Rightarrow d t=5 x^4 d x$
So,
$\int 5 x^4 \sqrt{x^5+1} d x =\int \sqrt{t} d t$
$= \frac{2}{3} t^{\frac{3}{2}}=\frac{2}{3}\left(x^5+1\right)^{\frac{3}{2}}$
So, $\int_{-1}^1 5 x^4 \sqrt{x^5+1} d x$
$=\frac{2}{3}\left(\left(x^5+1\right)^{\frac{3}{2}}\right)_{-1}^1$
$=\frac{2}{3}\left(\left(1^5+1\right)^{\frac{3}{2}}-\left((-1)^5+1\right)^{\frac{3}{2}}\right)$
$=\frac{2}{3}\left(2^{\frac{3}{2}}-(0)^{\frac{3}{2}}\right)=\frac{2}{3}(2 \sqrt{2}-0)$
$=\frac{4 \sqrt{2}}{3} \text { }$

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