What is the integral of e^xsqrt(6+e^x) ?

The integral of e^xsqrt(6+e^x) is 2/3 (6+e^x)^{3/2}+C

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integral of e^xsqrt(6+e^x)

\int e^{x} 6 + e^{x} d x

\frac{2}{3} (6 + e^{x})^{\frac{3}{2}} + C

Solution steps

\int e^{x} 6 + 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 = 6 + e^{x}

= \frac{2}{3} (6 + e^{x})^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (6 + e^{x})^{\frac{3}{2}} + C

(\frac{1}{2} 3 x)^{^{'}}

\int \frac{50}{w ^{2} 25 - w ^{2}} d w

\int 4 x + 1 d x

\int θ tan^{2} (θ) d θ

2^{x} + x^{2} e^{x^{2}}

What is the integral of e^xsqrt(6+e^x) ?
The integral of e^xsqrt(6+e^x) is 2/3 (6+e^x)^{3/2}+C