# Substitution Integral

3.141592654

## Homework Statement

$$\int(x^5\sqrt{x^2+4})dx$$

The answer is given as: $$=105(x^2+4)^\frac{3}{2}(15x^4-48x^2+128)+C$$

## The Attempt at a Solution

$$u=\sqrt{x^2+4}$$

$$u^2=x^2+4$$

$$2udu=2xdx$$

$$udu=xdx$$

$$u^2-4=x^2$$

$$\int(x^5\sqrt{x^2+4})dx = \int(u^2+4)^2u^2du = \int(u^4-8u^2+16)u^2du$$

$$=\int(u^6-8u^4+16u^2)du = \int(u^6)du-8\int(u^4)du+16\int(u^2)du$$

$$=\frac{u^7}{7}-8\frac{u^5}{5}+16\frac{u^3}{3}+C$$

$$=\frac{(x^2+4)^\frac{7}{2}}{7}-8\frac{(x^2+4)^\frac{5}{2}}{5}+16\frac{(x^2+4)^\frac{3}{2}}{3}+C$$

$$=(x^2+4)^\frac{3}{2}[\frac{(x^2+4)^\frac{4}{2}}{7}-8\frac{(x^2+4)^\frac{2}{2}}{5}+\frac{16}{3}]+C$$

$$=(x^2+4)^\frac{3}{2}[\frac{1}{7}(x^2+4)^2-\frac{8}{5}(x^2+4)+\frac{16}{3}]+C$$

$$=(x^2+4)^\frac{3}{2}[\frac{1}{7}(x^4-8x^2+16)-\frac{8}{5}(x^2+4)+\frac{16}{3}]+C$$

$$=(x^2+4)^\frac{3}{2}[\frac{1}{7}x^4-\frac{8}{7}x^2+\frac{16}{7}+\frac{8}{5}x^2+\frac{32}{5}+\frac{16}{3}]+C$$

$$=(x^2+4)^\frac{3}{2}(\frac{1}{7}x^4+\frac{16}{35}x^2+\frac{1248}{105})+C$$

I'm obviously off by a bit from the stated answer, but I can't find where I may have went wrong.

$$\int(x^5\sqrt{x^2+4})dx = \int(u^2+4)^2u^2du = \int(u^4-8u^2+16)u^2du$$
$$=\frac{(x^2+4)^\frac{7}{2}}{7}-8\frac{(x^2+4)^\frac{5}{2}}{5}+16\frac{(x^2+4)^\fr ac{3}{2}}{3}+C$$