How do you integrate  x^3((x^4)+3)^5 dx?

May 3, 2016

${\left({x}^{4} + 3\right)}^{6} / 24 + C$

Explanation:

We have the integral

$\int {x}^{3} {\left({x}^{4} + 3\right)}^{5} \mathrm{dx}$

This can be solved fairly simply through substitution: we don't have to go through the hassle of expanding the binomial and then integrating term by term.

Let $u = {x}^{4} + 3$. This implies that $\mathrm{du} = 4 {x}^{3} \mathrm{dx}$. Since we only have ${x}^{3} \mathrm{dx}$ in the integral and not $4 {x}^{3} \mathrm{dx}$, multiply the interior of the integral by $4$ and the exterior by $1 / 4$ to balance this.

$\int {x}^{3} {\left({x}^{4} + 3\right)}^{5} \mathrm{dx} = \frac{1}{4} \int {\left({x}^{4} + 3\right)}^{5} \cdot 4 {x}^{3} \mathrm{dx} = \frac{1}{4} \int {u}^{5} \mathrm{du}$

Integrating this results in $\left(\frac{1}{4}\right) {u}^{6} / 6 + C$, and since $u = {x}^{4} + 3$, this becomes ${\left({x}^{4} + 3\right)}^{6} / 24 + C$.