What is the integral of x^4e^{1-x^5} ?

The integral of x^4e^{1-x^5} is -1/5 e^{1-x^5}+C

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integral of x^4e^{1-x^5}

\int x^{4} e^{1 - x^{5}} d x

- \frac{1}{5} e^{1 - x^{5}} + C

Solution steps

\int x^{4} e^{1 - x^{5}} d x

Apply u-substitution

: \int - \frac{e ^{u}}{5} d u

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

Take the constant out:

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

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

Use the common integral:

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

= - \frac{1}{5} e^{u}

Substitute back

u = 1 - x^{5}

= - \frac{1}{5} e^{1 - x^{5}}

Add a constant to the solution

= - \frac{1}{5} e^{1 - x^{5}} + C

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x \to a lim (3 x^{2} - 6 x)

\frac{d}{d x} (x^{3} - x^{2} - 3 x + 10)

What is the integral of x^4e^{1-x^5} ?
The integral of x^4e^{1-x^5} is -1/5 e^{1-x^5}+C