The Integral$ \int_{1}^{\infty}\frac{e^{x}+e^{3x}}{e^{x}-e^{5x}}dx$ is convergent, but Im getting divergent

Question

$$\int_{1}^{\infty}\frac{e^{x}+e^{3x}}{e^{x}-e^{5x}}dx$$

Marking e^x as t I get: $$\int \frac{t+t^{3}}{t-t^{5}}$$

Dividing this integral: $$\int 1+\int \frac{-1}{t^{4}}+\int t^{2}+\int \frac{-1}{t^2}$$ which in term equals to:$$t-\frac{1}{3t^{3}}+\frac{t^{3}}{3}-\frac{1}{t}$$

It's clear that after substituing infinity and 1 I get infinity, but the answer is $$-1+\frac{1}{2}\ln (e^{2}-1)$$

Where is my mistake?

Answer 1

Since $dx=dt/t$, you need to divide the whole integrand by $t$.

Answer 2

There are a ton of mistakes here, unfortunately. The key issue is that you've got something like

$$\frac{t + t^3}{t - t^5} = 1 + t^2 - \frac{1}{t^4} - \frac{1}{t^2}$$

where you've just mixed-and-matched all four terms. This is a (very) incorrect manipulation of the fractions. One way that you can tell the two sides are unrelated is that the left hand side tends to $1$ as $t \to \infty$, while the right hand side blows up.

The second and third issues, as pointed out in the other answers, are that you're missing $dt/t = dx$ from the substitution, and that you didn't change the bounds to $[e, \infty)$.

Answer 3

If you make the substitution $\;t=\mathrm e^x\iff x=\ln t$, so that $\;\mathrm dx=\dfrac{\mathrm d t}t$, we obtain $$\int_{1}^{\infty}\frac{\mathrm e^{x}+\mathrm e^{3x}}{\mathrm e^{x}-\mathrm e^{5x}}\,\mathrm dx=\int_{\mathrm e}^{\infty}\frac{t+t^3}{t-t^5}\dfrac{\mathrm d t}t = \int_{\mathrm e}^{\infty}\frac{1+t^2}{t-t^5}\,\mathrm d t.$$ Now you can prove the convergence using the comparison test: near $+\infty$ the integrand has a simple equivalent: $$\frac{1+t^2}{t-t^5}\sim_{+\infty}\frac{t^2}{-t^5}=-\frac1{t^3},$$ which has a convergent integral on the same inteval.

Answer 4

I think it should be: $$I=\int_1^\infty\frac{e^x+e^{3x}}{e^x-e^{5x}}dx$$ $u=e^x$ so $dx=\frac{du}{u}$ so: $$I=\int_e^\infty\frac{u+u^3}{u-u^5}\frac{1}{u}du=\int_1^\infty\frac{1+u^2}{u(1-u^4)}du=\int_1^\infty\frac{1+u^2}{u(1-u^2)(1+u^2)}du=\int_1^\infty\frac{1}{u(1-u^2)}du$$ $v=1-u^2$ $$I=-\frac{1}{2}\int_0^{-\infty}\frac{1}{v(1-v)}dv=\frac{1}{2}\int_{-\infty}^0\frac{1}{v(1-v)}dv$$ and this can easily be solved using partial fractions