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A bracket cubed

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Compute $\int x (x^{2} + 1)^{3} d x$ with lower limit $0$ and upper limit $2$ ; the result is a whole number.

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Hint

The factor in front is, up to a constant, the derivative of what sits inside the bracket.

Solution

Substitute $u = x^{2} + 1$ , so $d u = 2 x d x$ and $x d x = \frac{d u}{2}$ .
The limits become $u = 1$ and $u = 4 + 1 = 5$ .
The integral turns into $(\frac{1}{2})$ times the integral of $u^{3}$ from $1$ to $5$ , which is $\frac{5 ^{4} - 1}{8}$ .
Since $5^{4} = 625$ , it equals $\frac{625 - 1}{8} = 78$ .

Answer: $78$