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? 
 A: 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?
A: 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
A: HINT
Let's integrate by parts.
with:
$$f(t)={(2t+1)e^{2t}}$$
$$g'(t)=\frac{1}{(t+1)^2}\implies g(t)=-\frac{1}{(t+1)}$$
A: 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.
