About the integral $\int _{\ln 2} ^{\ln 3} \frac { e ^x} {e ^{4x}-1} \,d x$ How do I compute $$\int \limits _{\ln 2} ^{\ln 3} \frac {e ^x} {e ^{4x}-1} \,  d x $$
? Thank you so much for your answer.
 A: Put $e^x=y$ , the limits change to 2 & 3.
Now ,  $\displaystyle I=\int_{2}^{3} \dfrac{dy}{y^4-1} = \frac{1}{2} (\int_{2}^{3} \frac{dy}{y^2-1} - \frac{dy}{y^2+1}) = \frac{1}{4}( \int_{2}^{3} \frac{dy}{y-1}-\frac{dy}{y+1}) - \frac{1}{2} [\tan^{-1}(y)]_{2}^{3} = \frac{1}{4}(\ln2 -\ln1 - \ln4 + \ln3) - \frac{1}{2}(\tan^{-1}(3)-\tan^{-1}(2)) $
So , I hope it demonstrates the process clearly. 
A: $$I=\int_{\log 2}^{\log 3}\frac{e^{x}}{e^{4x}-1}\,dx = \int_{2}^{3}\frac{du}{u^4-1}\,du=\int_{2}^{3}\frac{1}{2}\left(\frac{1}{u^2-1}-\frac{1}{u^2+1}\right)\,du $$
leads to:
$$ I = \int_{2}^{3}\frac{1}{2}\left(\frac{1/2}{u-1}-\frac{1/2}{u+1}-\frac{1}{u^2+1}\right)\,du = \frac{1}{4}\log\left(\frac{3}{2}\right)-\frac{\arctan 3-\arctan 2}{2}$$
or to:
$$ I = \frac{1}{4}\left(\log\frac{3}{2}-\arctan\frac{7}{24}\right).$$
A: $$\int_{\ln(2)}^{\ln(3)}\frac{e^x}{e^{4x}-1}\space\text{d}x=$$

Substitute $u=e^x$ and $\text{d}u=e^x\space\text{d}x$.
This gives a new lower bound $u=e^{\ln(2)}=2$ and upper bound $u=e^{\ln(3)}=3$:

$$\int_{2}^{3}\frac{1}{u^4-1}\space\text{d}u=$$

Use partial fractions:

$$\int_{2}^{3}\left[\frac{1}{4(u-1)}-\frac{1}{4(u+1)}-\frac{1}{2(u^2+1)}\right]\space\text{d}u=$$
$$\frac{1}{4}\int_{2}^{3}\frac{1}{u-1}\space\text{d}u-\frac{1}{4}\int_{2}^{3}\frac{1}{u+1}\space\text{d}u-\frac{1}{2}\int_{2}^{3}\frac{1}{u^2+1}\space\text{d}u=$$
$$\frac{1}{4}\int_{2}^{3}\frac{1}{u-1}\space\text{d}u-\frac{1}{4}\int_{2}^{3}\frac{1}{u+1}\space\text{d}u-\frac{1}{2}\left[\arctan(u)\right]_2^3=$$

For the integral $\int\frac{1}{u+1}\space\text{d}u$:
Substitute $s=u+1$ and $\text{d}s=\text{d}u$.
This gives a new lower bound $s=2+1=3$ and upper bound $s=3+1=4$:

$$\frac{1}{4}\int_{2}^{3}\frac{1}{u-1}\space\text{d}u-\frac{1}{4}\int_{3}^{4}\frac{1}{s}\space\text{d}s-\frac{1}{2}\left[\arctan(u)\right]_2^3=$$
$$\frac{1}{4}\int_{2}^{3}\frac{1}{u-1}\space\text{d}u-\frac{1}{4}\left[\ln|s|\right]_3^4-\frac{1}{2}\left[\arctan(u)\right]_2^3=$$

Substitute $p=u-1$ and $\text{d}p=\text{d}u$.
This gives a new lower bound $p=2-1=1$ and upper bound $p=3-1=2$:

$$\frac{1}{4}\int_{1}^{2}\frac{1}{p}\space\text{d}p-\frac{1}{4}\left[\ln|s|\right]_3^4-\frac{1}{2}\left[\arctan(u)\right]_2^3=$$
$$\frac{1}{4}\left[\ln|p|\right]_1^2-\frac{1}{4}\left[\ln|s|\right]_3^4-\frac{1}{2}\left[\arctan(u)\right]_2^3=$$
$$\frac{1}{4}\left(\ln|2|-\ln|1|\right)-\frac{1}{4}\left(\ln|4|-\ln|3|\right)-\frac{1}{2}\left(\arctan(3)-\arctan(2)\right)=$$
$$\frac{1}{4}\left(\ln|2|-0\right)-\frac{1}{4}\left(\ln|4|-\ln|3|\right)-\frac{1}{2}\left(\arctan(3)-\arctan(2)\right)=$$
$$\frac{\ln(2)-\ln(4)+\ln(3)}{4}-\frac{\arctan(3)-\arctan(2)}{2}\approx0.03041774972495909$$
