Integral of exponential rational function

Question

I'm asked to find $$\int_0^{\ln 2}{e^{2x}\over{e^{4x}+3}} \text{ d}x$$ I can't for the life of me figure out how to integrate this.

Answer 1

Let $u=e^{2x}$. Then $du=2e^{2x}\,dx$ and $$\int_0^{\ln 2}\frac{e^{2x}}{e^{4x}+3}dx=\frac12\int_0^4\frac{1}{u^2+3}du$$

You can then use trig substitution.

Answer 2

Let us look at $$\int{e^{2x}\over{e^{4x}+3}}dx$$ Let $u=2x\quad du=2dx$ $$\int{e^{2x}\over{e^{4x}+3}}dx=\dfrac{1}{2}\int{\dfrac{e^u}{e^{2u}+3}}du$$ Let $v=e^u \quad dv=e^u du=vdu$ $$=\dfrac{1}{2}\int\dfrac{1}{e^{2ln(v)}+3}dv=\dfrac{1}{2}\int\dfrac{1}{v^2+3}dv$$ This seems like the inverse trig derivative (arctan) $$=\dfrac{1}{2}\int\dfrac{1}{3\left(\left(\frac{v}{\sqrt{3}}\right)^2+1\right)}$$ Let $w=\dfrac{v}{\sqrt{3}} \quad dw=\dfrac{1}{\sqrt{3}}dv$ $$\dfrac{\sqrt{3}}{6}\int\dfrac{1}{1+w^2}=\dfrac{\sqrt{3}}{6}\arctan(w)$$

You substitute for $w,v,u$ and plug in the numbers for the bounds and you'll get your answer, so I will leave that up to you. I hope you followed the steps.