What is the integral of 3(x^4)/(e^{x^5)} ?

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

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integral of 3(x^4)/(e^{x^5)}

\int 3 \frac{x ^{4}}{e ^{x^{5}}} d x

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

Solution steps

\int 3 \cdot \frac{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

= 3 \cdot \int \frac{x ^{4}}{e ^{x^{5}}} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

= 3 \cdot \int x^{4} e^{- x^{5}} d x

Apply u-substitution

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

= 3 \cdot \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

= 3 (- \frac{1}{5} \cdot \int e^{u} d u)

Use the common integral:

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

= 3 (- \frac{1}{5} e^{u})

Substitute back

u = - x^{5}

= 3 (- \frac{1}{5} e^{- x^{5}})

Simplify

3 (- \frac{1}{5} e^{- x^{5}}) : - \frac{3}{5} e^{- x^{5}}

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

Add a constant to the solution

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

\int_{0}^{\frac{π}{4}} 1

9 x + 1 - x^{2}

x \to 1 lim (\frac{x ^{2} - 3 x}{x - 1})

y = (x^{2} + 4 x - 3)^{4}

\int \frac{e ^{2 y}}{e ^{y} - 1} d y

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