Evaluate $\displaystyle \int{x^3\sqrt{x^2+36}} dx$
Stuck on this one.
i hope it can help you :
$\displaystyle u=x^2+36\implies\tfrac{1}{2}\,du=x\,dx$
$\displaystyle x^2=u-36$
$\displaystyle \int x^3\sqrt{x^2+36}\,dx=\int x^2\sqrt{x^2+36}\,xdx=\int (u-36)\sqrt{u}\,\tfrac{1}{2}du$$\displaystyle =\tfrac{1}{2}\int (u-36)\sqrt{u}\,du$
Anyone mind checking the complete solution before I submit it?
$\displaystyle \frac{1}{2}\int(u-36)\sqrt{u} = \frac{1}{2}\int{u^{\frac{3}{2}}-36u^{\frac{1}{2}}} $
$\displaystyle = \frac{1}{2}(\frac{2}{5}u^{\frac{5}{2}}-24u^{\frac{3}{2}})+ C = \frac{2}{10}(x^2+36)^{\frac{5}{2}} - 12(x^2+36)^{\frac{3}{2}} + C$
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