What is the integral of-5x^4e^{x^5} ?

The integral of-5x^4e^{x^5} is -e^{x^5}+C

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integral of-5x^4e^{x^5}

\int - 5 x^{4} e^{x^{5}} d x

- e^{x^{5}} + C

Solution steps

\int - 5 x^{4} e^{x^{5}} d x

Take the constant out:

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

= - 5 \cdot \int x^{4} e^{x^{5}} d x

Apply u-substitution

= - 5 \cdot \int \frac{e ^{u^{\frac{5}{4}}} u ^{\frac{1}{4}}}{4} d u

Take the constant out:

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

= - 5 \cdot \frac{1}{4} \cdot \int e^{u^{\frac{5}{4}}} u^{\frac{1}{4}} d u

Apply u-substitution

= - 5 \cdot \frac{1}{4} \cdot \int \frac{4 e ^{v}}{5} d v

Take the constant out:

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

= - 5 \cdot \frac{1}{4} \cdot \frac{4}{5} \cdot \int e^{v} d v

Use the common integral:

\int e^{v} d v = e^{v}

= - 5 \cdot \frac{1}{4} \cdot \frac{4}{5} e^{v}

Substitute back

= - 5 \cdot \frac{1}{4} \cdot \frac{4}{5} e^{(x^{4})^{\frac{5}{4}}}

Simplify

- 5 \cdot \frac{1}{4} \cdot \frac{4}{5} e^{(x^{4})^{\frac{5}{4}}} : - e^{x^{5}}

= - e^{x^{5}}

Add a constant to the solution

= - e^{x^{5}} + C

d er i v a t i v e f (x) = \frac{x}{32} (x + 3 x)

\frac{\partial}{\partial y} (sin (x^{2} y))

\frac{d}{d x} ((x^{3} + 6 x)^{4})

in v erse l a pl a ce \frac{2 e ^{- 2 s}}{3 s ^{2} + 12}

3 y^{^{'}} + y = 6 t, y (0) = 3