Question

Integral of x(2x+3)^99 dx

Answer 1

IBP
https://answers.yahoo.com/question/index?qid=20100307231322AAJZmYV

Answer 2

Expand the brackets first.
Obviously that's the way to go.




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jk

Answer 3

Substitution: let \(u=2x+3\), so that \(x=\dfrac{u-3}{2}\) and \(\dfrac{du}{2}=dx\).
\[\large\int x(2x+3)^{99}~dx=\int\frac{u-3}{4}u^{99}~du=\int\left(\frac{u^{100}}{4}-\frac{3u^{99}}{4}\right)~du\]