How do you integrate by substitution $\int x^{2} {(x^{3} + 5)}^{4} d x$ $int x^2(x^3+5)^4 dx$ ?

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Answer 1 · 2016-10-31T19:49:15

$\int x^{2} {(x^{3} + 5)}^{4} d x = \frac{{(x^{3} + 5)}^{5}}{15} + C$ $int x^2(x^3+5)^4 dx =(x^3+5 )^5/15+C$

We want to find $I = \int x^{2} {(x^{3} + 5)}^{4} d x$ $I=int x^2(x^3+5)^4 dx$

Let $u = x^{3} + 5 \Rightarrow \frac{d u}{d x} = 3 x^{2}$ $u=x^3+5 => (du)/dx=3x^2$ , or $3 x^{2} \frac{d x}{d u} = 1$ $3x^2dx/(du)=1$

We can then rewrite $I$ $I$ as follows:

$I = \int x^{2} u^{4} d x$ $I=int x^2u^4 dx$
$:. I=1/3 int u^4 (3x^2)dx$
$:. I=1/3 int u^4 (3x^2)dx/(du)du$ (by the chain rule)

And using the above result we can now substitute to get:
$I=1/3 int u^4 (1) du$
$:. I=1/3 int u^4 du$
$:. I=1/3 u^5/5+C$
$:. I=(x^3+5 )^5/15+C$

How do you integrate by substitution int x^2(x^3+5)^4 dx∫x2(x3+5)4dxint x^2(x^3+5)^4 dx?