How do you use substitution to integrate $4 x {(x^{2} + 3)}^{4} d x$ $4x( x^2 + 3)^4 dx$ ?

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Answer 1 · 2018-03-29T03:00:55

$\frac{2}{5} {(x^{2} + 3)}^{5} + C$ $2/5(x^2+3)^5+C$

Substitute $x^{2} + 3 = u$ $x^2+3 = u$ . Then $2 x d x = d u$ $2xdx = du$ .

So

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

How do you use substitution to integrate 4x( x^2 + 3)^4 dx4x(x2+3)4dx4x( x^2 + 3)^4 dx?