- #1

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## Homework Statement

[tex]\int(x^5\sqrt{x^2+4})dx[/tex]

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

## Homework Equations

## The Attempt at a Solution

[tex]u=\sqrt{x^2+4}[/tex]

[tex]u^2=x^2+4[/tex]

[tex]2udu=2xdx[/tex]

[tex]udu=xdx[/tex]

[tex]u^2-4=x^2[/tex]

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

[tex]=\int(u^6-8u^4+16u^2)du = \int(u^6)du-8\int(u^4)du+16\int(u^2)du[/tex]

[tex]=\frac{u^7}{7}-8\frac{u^5}{5}+16\frac{u^3}{3}+C[/tex]

[tex]=\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[/tex]

[tex]=(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[/tex]

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

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

[tex]=(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[/tex]

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

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