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#