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

This an integration by parts problem. I am asking for assistance with my method. I was taught to use the LIATE (Logarithmic function, Inverse Trig Function, Algebraic Function, Trigonometric Function, and Exponential function) mnemonic to decide which function would be "u" and "dv" so as to be utilized in the following formula:

$\int udv= uv-\int v\ du$

u would be what ever first letter of a function which came first in the mnemonic and dv would be the latter.

Now this problem is not fitting into my instructions, and I am bit confused as to how to solve it. Please do not leave answers without a clear written explanation in clear English. You do not have to solve the problem. I just would like to know if my method is correct.

No disrespect intended.

zhw.
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BenCarson2016
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  • This looks like it might be a candidate for double-integration-by-parts... – abiessu Oct 10 '15 at 01:53
  • A double integration by parts? I was only taught "Repeated by Parts" and "Circular by Parts". Can you recommend a website or video that teaches the method? – BenCarson2016 Oct 10 '15 at 01:57
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    Let $u=xe^{2x}$ and $dv=dx/(1+2x)^2$. "Magic" will happen. I dislike the problem, if we change the $e^{2x}$ to $e^{3x}$ then we are in deep trouble. – André Nicolas Oct 10 '15 at 01:57
  • you'll need to first integrate by parts with $u=xe^{2x}$ and $\frac{dv}{dx}=\frac{1}{(1+2x)^2}$. Integrating $\frac{1}{(1+2x)^2}$ becomes really easy when you use the substitution $z=(1+2x)$. Finishing off the integration by parts alot of the new integral cancels out and you should be left with an easy to integrate function. – Andrew Brick Oct 10 '15 at 02:03
  • $du= (x2e^{2x}+e^{2x}$ or $e^{2x}(x2+1)$ I have no idea what the integral of $\int \frac{1}{(1+2x)^2}$ is. Should I transfer the $(1+2x)^2$ to the numerator and utilize this rule $\int x^n+ dx=\frac{x^{n+1}}{n+1}+C$? – BenCarson2016 Oct 10 '15 at 02:05
  • I prefer to say $du=(1+2x)e^{2x},dx$. To find $v$, let $y=1+2x$. Then $dx=(1/2),dy$, and we are integrating $\frac{1}{2}\cdot \frac{1}{y^2}$. We get $-\frac{1}{2y}$. So we can take $v=-\frac{1}{1+2x}$. I expect you can finish now. In the remaining integral there will be very nice cancellation. – André Nicolas Oct 10 '15 at 02:14
  • $\frac{1}{2}\int\frac{xe^{2x}}{y^2}$dy not sure how this is helpful. How does this help me calculate a dv value? – BenCarson2016 Oct 10 '15 at 02:33

3 Answers3

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Let $u=xe^{2x}$ and let $dv=\frac{dx}{(1+2x)^2}$. Then $du=(1+2x)e^{2x}\,dx$.

We need to find $v$. So we need to find $\int \frac{dx}{(1+2x)^2}$. Let $y=1+2x$. Then $dy=2\,dx$, and therefore $dx=\frac{dy}{2}$. Thus $$\int\frac{dx}{(1+2x)^2}=\int \frac{dy}{2y^2}=-\frac{1}{2y}+C.$$ We will take $C=0$. So we are taking $v=-\frac{1}{2(1+2x)}$.

Now using the integration by parts formula, we find that our integral is $uv-\int v\,du$. Thus we get $$-\frac{xe^{2x}}{2(1+2x)}+\int \frac{(1+2x)e^{2x}}{2(1+2x)}\,dx.$$ Note the cancellation. We need to find $\int \frac{e^{2x}}{2}\,dx$. By substitution or by inspection, the integral is $\frac{e^{2x}}{4}+C$. Putting things together, we get $$\frac{e^{2x}}{4}-\frac{xe^{2x}}{2(1+2x)}+C.$$ We can make this look a little nicer by using a common denominator $4(1+2x)$. We get $$e^{2x}\frac{(1+2x)-2x}{4(1+2x)}+C,$$ or equivalently $$\frac{e^{2x}}{4(1+2x)}+C.$$

Remark: Things "worked out" because the derivative of $xe^{2x}$ had a $1+2x$ "in it," giving us magic cancellation later. I don't like this problem, because if we replace $e^{2x}$ by $e^{3x}$, and leave everything else unchanged, we get an integral that cannot be solved using elementary functions.

André Nicolas
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Hint

Whenever there is $e^{g(x)}$ terms involved i always try making use of this property:

$\int e^{g(x)}[f'(x) + g'(x)f(x)] dx = f(x) e^{g(x)}$

This formula is actually the brother of one of the most commonly used formula

$\int e^{x} [f(x) + f'(x)]= e^{x}f(x)$

Here $\displaystyle f(x) = \frac{1}{2x+1} $

Sujith Sizon
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$$ \begin{aligned} \int \frac{x e^{2 x}}{(1+2 x)^{2}} d x &-\frac{1}{2}\int x e^{2 x} d\left(\frac{1}{1+2 x}\right)\\ =&-\frac{x e^{2 x}}{2(1+2 x)}+\frac{1}{2} \int \frac{2 x e^{2 x}+e^{2 x}}{1+2 x} d x \\ =&-\frac{x e^{2 x}}{2(1+2 x)}+\frac{1}{2} \int e^{2 x} d x \\ =&-\frac{x e^{2 x}}{2(1+2 x)}+\frac{1}{4} e^{2 x}+C \\ =& \frac{e^{2 x}}{4(1+2 x)}+C \end{aligned} $$

Lai
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