What is the integral of 8e^xsqrt(9+e^x) ?

The integral of 8e^xsqrt(9+e^x) is 16/3 (9+e^x)^{3/2}+C

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integral of 8e^xsqrt(9+e^x)

\int 8 e^{x} 9 + e^{x} d x

\frac{16}{3} (9 + e^{x})^{\frac{3}{2}} + C

Solution steps

\int 8 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

= 8 \cdot \int e^{x} 9 + e^{x} d x

Apply u-substitution

: \int u d u

= 8 \cdot \int u d u

Apply the Power Rule

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

= 8 \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 9 + e^{x}

= 8 \cdot \frac{2}{3} (9 + e^{x})^{\frac{3}{2}}

Simplify

8 \cdot \frac{2}{3} (9 + e^{x})^{\frac{3}{2}} : \frac{16}{3} (9 + e^{x})^{\frac{3}{2}}

= \frac{16}{3} (9 + e^{x})^{\frac{3}{2}}

Add a constant to the solution

= \frac{16}{3} (9 + e^{x})^{\frac{3}{2}} + C

f (x) = \frac{x ^{7}}{7} - \frac{7}{x ^{7}}, at x = - 1

4

\frac{d}{d x} (x sin (x) - cos (x))

\int 8 x^{3} \cdot (2 x^{4} + 4)^{4} d x

6 (ln (x) + 1)

What is the integral of 8e^xsqrt(9+e^x) ?
The integral of 8e^xsqrt(9+e^x) is 16/3 (9+e^x)^{3/2}+C