What is the integral of e^xsqrt(e^{x+1)} ?

The integral of e^xsqrt(e^{x+1)} is sqrt(e) 2/3 e^{(3x)/2}+C

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integral of e^xsqrt(e^{x+1)}

\int e^{x} e^{x + 1} d x

e \frac{2}{3} e^{\frac{3 x}{2}} + C

Solution steps

\int e^{x} e^{x + 1} d x

e^{x} e^{x + 1} = e^{1} e^{x} e^{x}

= \int e^{1} e^{x} e^{x} d x

Take the constant out:

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

= e^{1} \cdot \int e^{x} e^{x} d x

Simplify

= e \cdot \int e^{x} e^{x} d x

Apply u-substitution

: \int u d u

= e \cdot \int u d u

Apply the Power Rule

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

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

Substitute back

u = e^{x}

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

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

= e \frac{2}{3} e^{\frac{3 x}{2}}

Add a constant to the solution

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

x \to 6 lim (\frac{x - 6}{∣ x - 6 ∣})

n = 1 \sum \infty \frac{n - 1}{n + 1}

f (x) = 7 x e^{x} csc (x)

x y^{^{'}} - 3 y = 7 x

\frac{s}{( s ^{2} - 1 )}

What is the integral of e^xsqrt(e^{x+1)} ?
The integral of e^xsqrt(e^{x+1)} is sqrt(e) 2/3 e^{(3x)/2}+C