Find $\int (e^{2x}+e^{3x})^\frac{1}{2}dx$

Question

$$\int (e^{2x}+e^{3x})^\frac{1}{2}dx$$ I'm not sure what substitution I'm supposed to make here. Can someone help?

Answer 1

$$ \int \sqrt{e^{2x}+e^{3x}}dx = \int e^x\sqrt{1+e^x}dx $$ Let $ u \equiv e^x+1$. So we have $$ \int \sqrt{u}du = \frac{2\sqrt{e^x+1}^3}{3} + C $$

Answer 2

Hint: $$\int (e^{2x}+e^{3x})^\frac{1}{2}dx=\int e^x \sqrt{e^x+1} dx$$