1. ## Indefinite Integral

Evaluate $\int{x^3\sqrt{x^2+36}} dx$

Stuck on this one.

2. Originally Posted by Em Yeu Anh
Evaluate $\int{x^3\sqrt{x^2+36}}\,dx$

Stuck on this one.

Apply the substitution $u=x^2+36\implies\tfrac{1}{2}\,du=x\,dx$.

Thus, $\int x^3\sqrt{x^2+36}\,dx\xrightarrow{u=x^2+36}{}\tfrac {1}{2}\int (u-36)\sqrt{u}\,du$.

Can you take it from here?

3. Originally Posted by Chris L T521
Apply the substitution $u=x^2+36\implies\tfrac{1}{2}\,du=\,dx$.

Thus, $\int x^3\sqrt{x^2+36}\,dx\xrightarrow{u=x^2+36}{}\tfrac {1}{2}\int (u-36)\sqrt{u}\,du$.

Can you take it from here?
I'm slightly confused on the last part that has (u-36); is that supposed to represent the x^3?

$u=x^2+36\implies\tfrac{1}{2}\,du=x\,dx$

$x^2=u-36$

$\int x^3\sqrt{x^2+36}\,dx=\int x^2\sqrt{x^2+36}\,xdx=\int (u-36)\sqrt{u}\,\tfrac{1}{2}du$ $=\tfrac{1}{2}\int (u-36)\sqrt{u}\,du$

5. Anyone mind checking the complete solution before I submit it?

$\frac{1}{2}\int(u-36)\sqrt{u} = \frac{1}{2}\int{u^{\frac{3}{2}}-36u^{\frac{1}{2}}}$

$= \frac{1}{2}(\frac{2}{5}u^{\frac{5}{2}}-24u^{\frac{3}{2}})+ C = \frac{2}{10}(x^2+36)^{\frac{5}{2}} - 12(x^2+36)^{\frac{3}{2}} + C$

6. Originally Posted by Em Yeu Anh
Anyone mind checking the complete solution before I submit it?

$\frac{1}{2}\int(u-36)\sqrt{u} = \frac{1}{2}\int{u^{\frac{3}{2}}-36u^{\frac{1}{2}}}$

$= \frac{1}{2}(\frac{2}{5}u^{\frac{5}{2}}-24u^{\frac{3}{2}})+ C = \frac{2}{10}(x^2+36)^{\frac{5}{2}} - 12(x^2+36)^{\frac{3}{2}} + C$