What is the integral of e^xsqrt(41+e^x) ?

The integral of e^xsqrt(41+e^x) is 2/3 (41+e^x)^{3/2}+C

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integral of e^xsqrt(41+e^x)

\int e^{x} 41 + e^{x} d x

\frac{2}{3} (41 + e^{x})^{\frac{3}{2}} + C

Solution steps

\int e^{x} 41 + e^{x} d x

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 41 + e^{x}

= \frac{2}{3} (41 + e^{x})^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (41 + e^{x})^{\frac{3}{2}} + C

cos^{2} (e^{c o s^{2} (t)})

\frac{280}{( s ^{2} + 14 s + 245 )}

\frac{\partial}{\partial x} (4 x^{3})

f (x) = 8 x - x^{2}, at x = 1

\int (2 + 3 x^{2} - 8 x^{3}) d x

What is the integral of e^xsqrt(41+e^x) ?
The integral of e^xsqrt(41+e^x) is 2/3 (41+e^x)^{3/2}+C