What is the integral of (e^x-3)^4e^x ?

The integral of (e^x-3)^4e^x is ((e^x-3)^5)/5+C

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integral of (e^x-3)^4e^x

\int (e^{x} - 3)^{4} e^{x} d x

\frac{( e ^{x} - 3 ) ^{5}}{5} + C

Solution steps

\int (e^{x} - 3)^{4} e^{x} d x

Apply u-substitution

: \int (u - 3)^{4} d u

= \int (u - 3)^{4} d u

Apply u-substitution

: \int v^{4} d v

= \int v^{4} d v

Apply the Power Rule

: \frac{v ^{5}}{5}

= \frac{v ^{5}}{5}

Substitute back

= \frac{( e ^{x} - 3 ) ^{5}}{5}

Add a constant to the solution

= \frac{( e ^{x} - 3 ) ^{5}}{5} + C

\int_{2}^{3} 3 x 2 x^{2} - 1 d x

x \to 0 lim ((2 x + 1) (x - 3))

\frac{8}{x}

\frac{\partial}{\partial x} (vx)

y^{^{'}} - 2 y = - 3 e^{t}

What is the integral of (e^x-3)^4e^x ?
The integral of (e^x-3)^4e^x is ((e^x-3)^5)/5+C