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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

\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}}

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

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

Add a constant to the solution

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

\int 6 e^{2 x + 1} d x

\int 9 cos^{2} (t) d t

\int \frac{x + 4}{x ^{2} + 3 x + 2} d x

\int 3 t - 8 d t

\int \frac{2 x}{( x - 1 ) ( x ^{2} + 1 )} d x

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