Question

Calculus 2
can some one help me on integration by parts?

Answer 1

\[\int\limits_{}^{} xe^(2x) /(1+2x)^2\]

Answer 2

\[\int\frac{xe^{2x}}{(1+2x)^2}\,dx~~?\]

Answer 3

yes! sorry for the typo

Answer 4

No problem. What have you tried?

Answer 5

trying to figure out which one is my u so i can let that equal to du and then dv to get my v. i chose u=xe^2x and dv=1/(1+2x)^2

Answer 6

Typically, when you're given a polynomial or rational function times the exponential, i.e. something of the form
\[\int \frac{P(x)}{Q(x)}e^{ax}\,dx\]
(where \(P\) and \(Q\) are polynomials) the most efficient partition is taking \(u\) to be the exponential and \(dv\) to be the poly/rational.

In this case, I'd suggest
\[\begin{matrix}u=e^{2x}&&&dv=\frac{x}{(1+2x)^2}\,dx\end{matrix}\]

Answer 7

okay so that means i would have du=2e^2x and v=?

Answer 8

That doesn't mean your suggestion isn't right, though. See what happens when we use your setup:
\[\begin{matrix}
u=xe^{2x}&&&dv=\frac{dx}{(1+2x)^2}\\
du=e^{2x}(2x+1)\,dx&&&v=-\frac{1}{2(1+2x)}
\end{matrix}\]
and so
\[\int\frac{xe^{2x}}{(1+2x)^2}\,dx=-\frac{1}{2}\frac{xe^{2x}}{(1+2x)}+\frac{1}{2}\int\frac{e^{2x}\cancel{(1+2x)}}{\cancel{1+2x}}\,dx\]

Answer 9

Stick with your setup, it's actually simpler! Meanwhile, here's what mine gives:
\[\begin{matrix}u=e^{2x}&&&dv=\frac{x}{(1+2x)^2}\,dx\\
du=2e^{2x}\,dx&&&\left\{\begin{align*}v&=\int\frac{x}{(1+2x)^2}\,dx\\
&=\frac{1}{2}\int\frac{\frac{t-1}{2}}{t^2}\,dt&(t=1+2x)\\
&=\frac{1}{4}\int\frac{t-1}{t^2}\,dt\\
&=\frac{1}{4}\int\left(\frac{1}{t}-\frac{1}{t^2}\right)\,dt\\
&=\frac{1}{4}\left(\ln|t|+\frac{1}{t}\right)\\
&=\frac{1}{4}\ln|1+2x|+\frac{1}{4(1+2x)}\end{align*}\right.
\end{matrix}\]
This doesn't seem to end up with anything easy.

Answer 10

ok thank you