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

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Answer 1 · 2018-03-29T03:00:55

$2/5(x^2+3)^5+C$

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

So

$int 4x( x^2 + 3)^4 dx = int 2u^4 du = 2/5u^5+C$
$qquad = 2/5(x^2+3)^5+C$

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