# Substitution Integral

## 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.

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Mark44
Mentor
I don't see anything obviously wrong, other than a typo or two that didn't affect your calculuations.
$$\int(x^5\sqrt{x^2+4})dx = \int(u^2+4)^2u^2du = \int(u^4-8u^2+16)u^2du$$
The first integral in u should have (u^2 - 4), not (u^2 + 4), but the following expression is correct.

In the next line, you have your antiderivative. As a sanity check, take its derivative, and if you get the original integrand, your work is good.
$$=\frac{(x^2+4)^\frac{7}{2}}{7}-8\frac{(x^2+4)^\frac{5}{2}}{5}+16\frac{(x^2+4)^\fr ac{3}{2}}{3}+C$$

As a final check, if the derivative of your last expression is equal to the integrand, then your answer works. If that's the case, you might want to differentiate the answer you are given and see if it works, also.