What is the integral of e^xsqrt(a-be^x) ?

The integral of e^xsqrt(a-be^x) is -(2(a-be^x)^{3/2})/(3b)+C

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integral of e^xsqrt(a-be^x)

\int e^{x} a - b e^{x} d x

- \frac{2 ( a - b e ^{x} ) ^{\frac{3}{2}}}{3 b} + C

Solution steps

\int e^{x} a - b e^{x} d x

Apply u-substitution

: \int a - b u d u

= \int a - b u d u

Apply u-substitution

: \int - \frac{v}{b} d v

= \int - \frac{v}{b} d v

Take the constant out:

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

= - \frac{1}{b} \cdot \int v d v

Apply the Power Rule

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

= - \frac{1}{b} \cdot \frac{2}{3} v^{\frac{3}{2}}

Substitute back

= - \frac{1}{b} \cdot \frac{2}{3} (a - b e^{x})^{\frac{3}{2}}

Simplify

- \frac{1}{b} \cdot \frac{2}{3} (a - b e^{x})^{\frac{3}{2}} : - \frac{2 ( a - b e ^{x} ) ^{\frac{3}{2}}}{3 b}

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

Add a constant to the solution

= - \frac{2 ( a - b e ^{x} ) ^{\frac{3}{2}}}{3 b} + C

y^{^{''}} + y^{^{'}} + 2 y = 0

\frac{d}{d x} ((5 x^{2} - 9) (8 x + 3))

(9.7) (3)

x e^{4 x}

\frac{d y}{d x} = x^{2} - y

What is the integral of e^xsqrt(a-be^x) ?
The integral of e^xsqrt(a-be^x) is -(2(a-be^x)^{3/2})/(3b)+C