How do you integrate $int (2x+3)sin(x^2+3x)$ by integration by parts method?

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Answer 1 · 2017-02-08T07:22:22

The integrand is perfect for $u$ -substitution.

I think using I.B.P would get really messy; especially if you're familiar with the L.I.A.T.E mnemonic . . .

You'd end up having to find the antiderivative of $sin(x^2+3x)$ which is REALLY ugly.

Just by looking at the integrand, we should let:

$u=x^2+3x$
$du=2x+3dx$

$intsin(u)du$

$=-cos(u) +C$

Plug $u$ back in:

$=-cos(x^2+3x)+C$

How do you integrate int (2x+3)sin(x^2+3x)int (2x+3)sin(x^2+3x) by integration by parts method?