How do you find the antiderivative of $e^{2 x} \cdot \sqrt{e^{x} + 1}$ $e^(2x) * sqrt(e^x + 1)$ ?

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Answer 1 · 2016-09-18T20:47:51

$\frac{2}{15} {(e^{x} + 1)}^{\frac{3}{2}} (3 e^{x} - 2) + C$ $2/15(e^x+1)^(3/2)(3e^x-2)+C$ .

Let, $I = \int e^{2 x} \sqrt{e^{x} + 1} d x$ $I=inte^(2x)sqrt(e^x+1)dx$

We use the subst. $e^{x} + 1 = t^{2}, or, e^{x} = t^{2} - 1, so that, e^{x} d x = 2 t d t$ $e^x+1=t^2, or, e^x=t^2-1", so that, "e^xdx=2tdt$ .

$:. I=inte^xsqrt(e^x+1)*e^xdx$

$=int(t^2-1)sqrt(t^2)*2tdt$

$=2int(t^4-t^2)dt$

$=2(t^5/5-t^3/3)$

$=2/15(3t^5-5t^3)$

$=2/15t^3(3t^2-5)$

$=2/15(e^x+1)^(3/2)({3(e^x+1)-5}..............[as, t=(e^x+1)^(1/2)]$

$=2/15(e^x+1)^(3/2)(3e^x-2)+C$ .

How do you find the antiderivative of e^(2x) * sqrt(e^x + 1)e2x⋅√ex+1e^(2x) * sqrt(e^x + 1)?