Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

\[\int\limits_{}^{} 3x^2e^2x dx\]

i took \[u=X^2 dv=e^2x\]

\[du=2x dx v=\frac{ e^2x }{ 2 }\]

\[uv-\int\limits\limits_{}^{}v dv\]

\[(\frac{ 3 }{ 2 }e ^{2x} x^2 -\]

after that confused

how to integrate v

thx

can u show me how to integrate sin(x^3) i mean just tell me how to do i will to the rest

does this include series because i am seeing for the first time is this cal 2

tylors series

ya thx i will check that out but ^ how did u get (3rd step) 2xe^2x dx

That's
\[v\;du = \frac{1}{2}2xe^{2x}dx,\text{ with the $\frac{1}{2}$ factored out.}\]

i got that i mean it is \[\int\limits_{}^{}v \] so v=\[\frac{ 1 }{ 2 }e^2x\]

i dont get where u got 2x in front of e^2x (typo ^ it is e^(2x))

The 2x comes from the du term. Since u = x², we get du = 2x dx

ohh so it is \[\int\limits_{}^{}v du\]

Yep, integration by parts, using the u and dv substitutions, yields
\[uv-\int v\;du\]

okay

thx

You're welcome