Jul 18, 2013 Thread starter #1 Y Yankel Active member Jan 27, 2012 398 Hello I need to solve \[\int \sqrt{x^{3}+4}\cdot x^{5}dx\] using the substitution method. I did \[u=x^{3}+4\] but I got stuck with it. thanks!

Hello I need to solve \[\int \sqrt{x^{3}+4}\cdot x^{5}dx\] using the substitution method. I did \[u=x^{3}+4\] but I got stuck with it. thanks!

Jul 18, 2013 Admin #2 M MarkFL Administrator Staff member Feb 24, 2012 13,775 I would try integration by parts, where: \(\displaystyle u=x^3\,\therefore\,du=3x^2\,dx\) \(\displaystyle dv=x^2\sqrt{x^3+4}\,dx\, \therefore\,v=\frac{2}{9}\left(x^3+4 \right)^{\frac{3}{2}}\) Can you proceed?

I would try integration by parts, where: \(\displaystyle u=x^3\,\therefore\,du=3x^2\,dx\) \(\displaystyle dv=x^2\sqrt{x^3+4}\,dx\, \therefore\,v=\frac{2}{9}\left(x^3+4 \right)^{\frac{3}{2}}\) Can you proceed?

Jul 18, 2013 #3 A awkward Member Feb 18, 2012 36 Your original idea of substitution will also work. Letting $u = x^3+4$ we have $du = 3 x^2 dx$ $x^3 = u-4$ so $\int \sqrt{x^3+4} \; x^5 \;dx = \int \sqrt{x^3+4} \; x^3 \cdot x^2 \;dx = \frac{1}{3} \int \; \sqrt{u} (u-4) \; dx$ and you can probably take it from there...

Your original idea of substitution will also work. Letting $u = x^3+4$ we have $du = 3 x^2 dx$ $x^3 = u-4$ so $\int \sqrt{x^3+4} \; x^5 \;dx = \int \sqrt{x^3+4} \; x^3 \cdot x^2 \;dx = \frac{1}{3} \int \; \sqrt{u} (u-4) \; dx$ and you can probably take it from there...