# Integral of $\frac{(2t+1)e^{2t}}{(t+1)^2}$

$$\frac{(2t+1)e^{2t}}{(t+1)^2}$$ I came across this integral to solve a differential equation, but I have no clue how to actually integrate this. I know the answer should be $e^{2t}/(t+1)$. I have tried substitution and integration by parts, but neither seem to make the problem any easier. I have tried adding another $e^{2t}$ and subtracting it afterwards (like adding zero), but this too seemed to make the problem harder.

Is there a standard approach for this type of integral?

Well the (very) fast way would be observing that: $$\left( \frac{e^{2t}}{t+1} \right)' = \frac{\left( 2t+1 \right) e^{2t}}{\left( t+1 \right)^2}$$ but you may not see this or may not want to depend on being able to see this.

Let $u=t+1 \iff t = u-1$ to get: $$\int \frac{\left( 2t+1 \right) e^{2t}}{\left( t+1 \right)^2} \,\mbox{d}t = \int \frac{\left( 2u-1 \right) e^{2u-2}}{u^2} \,\mbox{d}u$$ Now you can continue with integration by parts. Does that help?

• I'll give this a shot, one moment please. Commented Dec 8, 2017 at 14:19
• Oh I see it now, I was struggling to find $\int\frac{1}{u^2}e^{2u-2}$, but this term cancels from integration by parts, so I don't need to actually be able to integrate it. Commented Dec 8, 2017 at 14:24
• Alright! Note that you can split off a (constant) factor and continue with a simpler exponent, if you want: $e^{2u-2} = e^{-2}e^{2u}$. Commented Dec 8, 2017 at 14:26
• I think that substitution is not necessary to proceed to integration by parts, I see this way as a double work.
– user
Commented Dec 8, 2017 at 15:03
• @gimusi It's not necessary, but unless you see the clever integration by parts doing it in one turn (as suggested in your answer - which is fine!), this paves the way for splitting the integral in two and performing ("standard", clearly visible) integration by parts on $u^{-1}e^{2u}$ and $u^{-2}e^{2u}$. Commented Dec 8, 2017 at 15:05

Setting $$t+1=u$$ then we get $$du=dt$$ and $$t=u-1$$ and our integral is given by $$\int\frac{(2u-1)e^{2(u-1)}}{u^2}du$$ can you proceed? and this integral can not expressed by the known elementary functions

• Shouldn't it be $(2u-1)$ in the numerator then? Commented Dec 8, 2017 at 14:18
• yes it should, corrected Commented Dec 8, 2017 at 14:21

Note that $f$ is a fraction with denominator a square: $(t+1)^2$. Looking at the numerator, we see something similar to $t+1$ (that is, $2t+1$) multiplied with $e^{2t}$. Both these are characteristics of the quotient rule: $$\left(\frac fg\right)'=\frac{f'g-fg'}{g^2}$$ So we may try as an ansatz $$\frac{e^{2t}}{t+1}$$ and differentiating this does give $f$. You may not be as lucky as this in general, though.

HINT

Let's integrate by parts.

with:

$$f(t)={(2t+1)e^{2t}}$$

$$g'(t)=\frac{1}{(t+1)^2}\implies g(t)=-\frac{1}{(t+1)}$$