What is the integral of 4e^xsqrt(10+e^x) ?

The integral of 4e^xsqrt(10+e^x) is 8/3 (10+e^x)^{3/2}+C

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integral of 4e^xsqrt(10+e^x)

\int 4 e^{x} 10 + e^{x} d x

\frac{8}{3} (10 + e^{x})^{\frac{3}{2}} + C

Solution steps

\int 4 e^{x} 10 + e^{x} d x

Take the constant out:

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

= 4 \cdot \int e^{x} 10 + e^{x} d x

Apply u-substitution

: \int u d u

= 4 \cdot \int u d u

Apply the Power Rule

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

= 4 \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 10 + e^{x}

= 4 \cdot \frac{2}{3} (10 + e^{x})^{\frac{3}{2}}

Simplify

4 \cdot \frac{2}{3} (10 + e^{x})^{\frac{3}{2}} : \frac{8}{3} (10 + e^{x})^{\frac{3}{2}}

= \frac{8}{3} (10 + e^{x})^{\frac{3}{2}}

Add a constant to the solution

= \frac{8}{3} (10 + e^{x})^{\frac{3}{2}} + C

\frac{1 - 5 t}{6 + t}

\frac{\partial}{\partial x} (x^{4} - a x^{2} + a^{2})

n = 1 \sum \infty n^{\frac{3}{2}}

t \cdot \frac{d y}{d t} = t^{3} + 11 t^{3} y

2 x^{5} + 3 x^{3} - 3

What is the integral of 4e^xsqrt(10+e^x) ?
The integral of 4e^xsqrt(10+e^x) is 8/3 (10+e^x)^{3/2}+C