What is the integral of 2e^xsqrt(9+e^x) ?

The integral of 2e^xsqrt(9+e^x) is 4/3 (9+e^x)^{3/2}+C

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integral of 2e^xsqrt(9+e^x)

\int 2 e^{x} 9 + e^{x} d x

\frac{4}{3} (9 + e^{x})^{\frac{3}{2}} + C

Solution steps

\int 2 e^{x} 9 + e^{x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int e^{x} 9 + e^{x} d x

Apply u-substitution

: \int u d u

= 2 \cdot \int u d u

Apply the Power Rule

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

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

Substitute back

u = 9 + e^{x}

= 2 \cdot \frac{2}{3} (9 + e^{x})^{\frac{3}{2}}

Simplify

2 \cdot \frac{2}{3} (9 + e^{x})^{\frac{3}{2}} : \frac{4}{3} (9 + e^{x})^{\frac{3}{2}}

= \frac{4}{3} (9 + e^{x})^{\frac{3}{2}}

Add a constant to the solution

= \frac{4}{3} (9 + e^{x})^{\frac{3}{2}} + C

\int (- 5 sec^{2} (x)) d x

\int_{- 2}^{- 1} x^{- 3} d x

x \to 1 lim (\frac{1 - \frac{1}{x}}{1 - x})

\int_{0}^{π} sec^{2} (\frac{t}{3}) d t

What is the integral of 2e^xsqrt(9+e^x) ?
The integral of 2e^xsqrt(9+e^x) is 4/3 (9+e^x)^{3/2}+C