How do you use substitution to integrate $x^{3} \sqrt{x - 3}$ $x^3sqrt(x-3)$ ?

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Answer 1 · 2015-10-15T03:16:27

See the explanation.

Let $u = x - 3$ $u = x-3$ so $d u = d x$ $du=dx$ and $x = u + 3$ $x = u+3$

Then $\int x^{3} \sqrt{x - 3} d x = \int {(u + 3)}^{3} u^{\frac{1}{2}} d u$ $intx^3sqrt(x-3) dx = int (u+3)^3u^(1/2) du$

Now expand ${(u + 3)}^{3}$ $(u+3)^3$ , distribute $u^{\frac{1}{2}}$ $u^(1/2)$ and finish by integrating term by term.

How do you use substitution to integrate x3√x−3x^3sqrt(x-3)?