What is the integral of 7e^xsqrt(2+e^x) ?

The integral of 7e^xsqrt(2+e^x) is 14/3 (2+e^x)^{3/2}+C

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integral of 7e^xsqrt(2+e^x)

\int 7 e^{x} 2 + e^{x} d x

\frac{14}{3} (2 + e^{x})^{\frac{3}{2}} + C

Solution steps

\int 7 e^{x} 2 + e^{x} d x

Take the constant out:

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

= 7 \cdot \int e^{x} 2 + e^{x} d x

Apply u-substitution

: \int u d u

= 7 \cdot \int u d u

Apply the Power Rule

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

= 7 \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 2 + e^{x}

= 7 \cdot \frac{2}{3} (2 + e^{x})^{\frac{3}{2}}

Simplify

7 \cdot \frac{2}{3} (2 + e^{x})^{\frac{3}{2}} : \frac{14}{3} (2 + e^{x})^{\frac{3}{2}}

= \frac{14}{3} (2 + e^{x})^{\frac{3}{2}}

Add a constant to the solution

= \frac{14}{3} (2 + e^{x})^{\frac{3}{2}} + C

\frac{d}{d x} (- \frac{1}{4} sin (\frac{x}{2}))

y = 2 x y^{^{'}}

(x + 1) (x^{3} + x - 2)

y^{^{'}} - \frac{6 y}{x} = (\frac{y}{x ^{2}})^{4}

\int \frac{( x + 3 ) ^{4}}{x} d x

What is the integral of 7e^xsqrt(2+e^x) ?
The integral of 7e^xsqrt(2+e^x) is 14/3 (2+e^x)^{3/2}+C