What is the integral of e^xsqrt(39+e^x) ?

The integral of e^xsqrt(39+e^x) is 2/3 (39+e^x)^{3/2}+C

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integral of e^xsqrt(39+e^x)

\int e^{x} 39 + e^{x} d x

\frac{2}{3} (39 + e^{x})^{\frac{3}{2}} + C

Solution steps

\int e^{x} 39 + 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 = 39 + e^{x}

= \frac{2}{3} (39 + e^{x})^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (39 + e^{x})^{\frac{3}{2}} + C

f (x) = x^{3} + 3, (1, 2)

\int - 3 x d x

e^{3 t}

\int \frac{1}{x ^{2} + 4 x} d x

y^{^{''}} - y^{^{'}} - 2 y = - 6 t + 6 t^{2}

What is the integral of e^xsqrt(39+e^x) ?
The integral of e^xsqrt(39+e^x) is 2/3 (39+e^x)^{3/2}+C