How do you find the integral of $e^{2 x} \sqrt{1 + e^{2 x}} d x$ $e^(2x) sqrt(1 + e^(2x)) dx$ ?

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Answer 1 · 2015-06-09T10:36:04

$\frac{1}{3} {(1 + e^{2 x})}^{\frac{3}{2}}$ $1/3 (1+e^(2x))^(3/2)$ +C

Let u= $1 + e^{2 x}$ $1+e^(2x)$ , du= 2 $e^{2 x}$ $e^(2x)$ dx

Accordingly, $\int e^{2 x} \sqrt{1 + e^{2 x} d x}$ $int e^(2x) sqrt(1+e^(2x) dx$ = $\frac{1}{2} \int u^{\frac{1}{2}} d u$ $1/2int u^(1/2) du$

= $\frac{1}{3} u^{\frac{3}{2}}$ $1/3 u^(3/2)$ +C

= $\frac{1}{3} {(1 + e^{2 x})}^{\frac{3}{2}}$ $1/3 (1+e^(2x))^(3/2)$ +C

How do you find the integral of e^(2x) sqrt(1 + e^(2x)) dxe2x√1+e2xdxe^(2x) sqrt(1 + e^(2x)) dx?