What is the integral of (2x^3)/(e^{x^4)} ?

The integral of (2x^3)/(e^{x^4)} is -1/2 e^{-x^4}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x^3)/(e^{x^4)}

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

- \frac{1}{2} e^{- x^{4}} + C

Solution steps

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

Take the constant out:

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

= 2 \cdot \int \frac{x ^{3}}{e ^{x^{4}}} d x

Apply exponent rule:

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

\frac{x ^{3}}{e ^{x^{4}}} = (x^{3}) e^{- x^{4}}

= 2 \cdot \int x^{3} e^{- x^{4}} d x

Apply u-substitution

: \int - \frac{e ^{u}}{4} d u

= 2 \cdot \int - \frac{e ^{u}}{4} d u

Take the constant out:

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

= 2 (- \frac{1}{4} \cdot \int e^{u} d u)

Use the common integral:

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

= 2 (- \frac{1}{4} e^{u})

Substitute back

u = - x^{4}

= 2 (- \frac{1}{4} e^{- x^{4}})

Simplify

2 (- \frac{1}{4} e^{- x^{4}}) : - \frac{1}{2} e^{- x^{4}}

= - \frac{1}{2} e^{- x^{4}}

Add a constant to the solution

= - \frac{1}{2} e^{- x^{4}} + C

\int \frac{1}{( x ^{2} + a ) \frac{3}{2}} d x

\int 3^{x} cos (2 x) d x

\int \frac{2 x}{x ^{2} - 5} d x

\int \frac{sin ( θ ) cos ( θ )}{3 + cos ^{2} ( θ )} d θ

\int \frac{4 x + 10}{( x - 2 ) ( x - 1 ) ^{2}} d x

What is the integral of (2x^3)/(e^{x^4)} ?
The integral of (2x^3)/(e^{x^4)} is -1/2 e^{-x^4}+C