# How to calculate the integral of this fraction

So I have this fairly complicated fraction, and I am trying to find its antiderivative.

$$\int \frac{7xe^{2x}}{(1+2x)^2} dx$$

I am not sure where to start. I was thinking using integration by parts, but then I tried substituting it with $u$'s:

$$u = 1+2x, du = 2 dx \to dx=\frac{du}{2} \\ = \int \frac{\frac{7u-7}{2}e^{u-1}}{u^2}\color{red}{\frac{ du}{2} }\\ = \frac{7}{4}\int \frac{e^{u-1}}{u} du - \frac{7}{4}\int \frac{e^{u-1}}{u^2} du$$

But now I am stuck and I can't go any further. Can someone please guide me through this problematic integration? Thanks!

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$$= \frac{7e^{-1}}{4}\bigg\{\color{red}{\int \frac{e^{u}}{u} du} - \int \frac{e^{u}}{u^2} du\bigg\}\\ = \frac{7e^{-1}}{4}\bigg\{\color{red}{\frac{e^{u}}{u}-\int \frac{-e^{u}}{u^2} du }- \int \frac{e^{u}}{u^2} du+c\bigg\}\ \text{integration by parts}\\ =\frac{7e^{-1}}{4}\frac{e^{u}}{u}+c =\frac{7e^{u-1}}{4u}+c =\frac{7e^{2x+1-1}}{4(2x+1)}+c =\frac{7e^{2x}}{4(2x+1)}+c$$

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Thanks for the response but I am having hard time figuring out what did you substitute for $u$. Can you explain it more? –  Derek 朕會功夫 Jan 31 at 5:26
i will in the answer now –  Semsem Jan 31 at 5:26
i use integration by parts in the red term –  Semsem Jan 31 at 5:29
$\int fg'dx=fg-\int gf'dx$ –  Semsem Jan 31 at 5:31
He started from where you left the problem. So $u=2x+1$. –  John Habert Jan 31 at 5:32

First, let's pull out the constant factor $7$. The denominator being a square suggests trying to find functions $f(x)$, $g(x)$ such that $$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2} = \frac{xe^{2x}}{(1+2x)^2}.$$ This in turn suggests trying $g(x) = 1+2x$, so that the numerator becomes $$(1+2x)f'(x)- 2f(x) = xe^{2x}.$$ If $f(x) = h(x) e^{2x}$ for some polynomial $h$, then $f'(x) = h'(x)e^{2x} + 2h(x) e^{2x} = (h'(x)+2h(x))e^{2x}$, hence $$xe^{2x} = (4xh(x) + (1+2x)h'(x))e^{2x}.$$ This is easily satisfied with the choice $h(x) = 1/4$, so $f(x) = e^{2x}/4$ and an antiderivative is $$7 \frac{f(x)}{g(x)} + C = \frac{7e^{2x}}{4(1+2x)} + C.$$ Of course, this method doesn't always work, but it is an interesting approach.

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It is interesting indeed I did not even think of ever trying this –  Millardo Peacecraft Jan 31 at 5:30
How did you get $xe^{2x} = (4xh(x) + (1+2x)h'(x))e^{2x}$ from the above? I was following but this line confuses me. –  Derek 朕會功夫 Jan 31 at 5:40
If $f(x) = h(x) e^{2x}$, then $f'(x) = (h'(x) + 2h(x))e^{2x}$, and $(1+2x)f'(x) - 2f(x) = (4xh(x) + (1+2x)h'(x))e^{2x}$, which we equate with the numerator of the integrand, assuming that it is the derivative of some quotient of functions. –  heropup Jan 31 at 5:43

$$\frac{d(e^{2x}f(x))}{dx}=e^{2x}f'(x)+2e^{2x}f(x)=e^{2x}[f'(x)+2f(x)]$$

If we choose $\displaystyle f'(x)=\frac{2A}{(1+2x)^2}, f(x)=-\frac A{1+2x}$

$$\frac{d\left[e^{2x}\cdot\frac{-A}{1+2x}\right]}{dx}=e^{2x}\left[\frac{2A}{(1+2x)^2}-2\frac A{1+2x}\right]=2Ae^{2x}\cdot \frac{1-(1+2x)}{(1+2x)^2}= \frac{-4Ax e^{2x}}{(1+2x)^2}$$

Compare with the given integral to find $-4A=7$

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d/dx(7*e^2/4*(2x+1))=7xe^2/x(1+2x)^2 =>d(7*e^2/4*(2x+1))=(7xe^2/x(1+2x)^2)dx Hence you get the required integral.

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