Integrate: $\int \frac {e^{6x}}{\sqrt {1-e^{3x}}} dx$

Question

Integrate: $\int \dfrac {e^{6x}}{\sqrt {1-e^{3x}}} dx$

My Attempt: $$=\int \dfrac {e^{6x}}{\sqrt {1-e^{3x}}} dx$$ Let $e^x = t$ $$dt=e^x.dx$$ Now, $$=\int \dfrac {t^5}{\sqrt {1-t^3}} dt$$

Answer 1

Hint: A far better substitution might by

$$t = 1 - e^{3x}$$

which entirely deals with the radical. This is motivated by the fact that the constant is immaterial after differentiating.

Answer 2

let $u = e^{3x}$.

Your integral becomes

$$ \frac{1}{3}\int \frac{u}{\sqrt{1-u}} du = $$ Now letting $u = v +1$

Your integral becomes $$ \frac{1}{3}\int \frac{v}{\sqrt{v}}+\frac{1}{\sqrt{v}} du $$

Answer 3

Hint

Consider $$I=\int \dfrac {e^{6x}}{\sqrt {1-e^{3x}}}\,dx$$ To get rid of the radical, let $$\sqrt {1-e^{3x}}=t^2 \implies x=\frac{1}{3} \log \left(1-t^4\right)\implies dx=-\frac{4 t^3}{3 \left(1-t^4\right)}\,dt$$ making $$I=\frac{4}{3} \int t \left(t^4-1\right)\,dt$$

Answer 4

Just for fun using differentials directly:

$\int \frac{e^{6x}}{\sqrt{1-e^{3x}}}dx=\int \frac{e^{3x}e^{3x}}{\sqrt{1-e^{3x}}}dx =\frac{1}{3}\int\frac{1-e^{3x}-1}{\sqrt{1-e^{3x}}}d(1-e^{3x})=$ $\frac{1}{3}\int\sqrt{1-e^{3x}}d(1-e^{3x})-\frac{1}{3}\int\frac{1}{\sqrt{1-e^{3x}}}d(1-e^{3x})=\frac{2}{9}(1-e^{3x})\sqrt{1-e^{3x}}-\frac{2}{3}\sqrt{1-e^{3x}}= -\frac{2}{9}\sqrt{1-e^{3x}}(2+e^{3x})$