What is the integral of (5x^2)/(e^{2x^3)} ?

The integral of (5x^2)/(e^{2x^3)} is -5/6 e^{-2x^3}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (5x^2)/(e^{2x^3)}

\int \frac{5 x ^{2}}{e ^{2 x^{3}}} d x

- \frac{5}{6} e^{- 2 x^{3}} + C

Solution steps

\int \frac{5 x ^{2}}{e ^{2 x^{3}}} d x

Take the constant out:

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

= 5 \cdot \int \frac{x ^{2}}{e ^{2 x^{3}}} d x

Apply exponent rule:

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

\frac{x ^{2}}{e ^{2 x^{3}}} = (x^{2}) e^{- 2 x^{3}}

= 5 \cdot \int x^{2} e^{- 2 x^{3}} d x

Apply u-substitution

: \int - \frac{e ^{u}}{6} d u

= 5 \cdot \int - \frac{e ^{u}}{6} d u

Take the constant out:

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

= 5 (- \frac{1}{6} \cdot \int e^{u} d u)

Use the common integral:

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

= 5 (- \frac{1}{6} e^{u})

Substitute back

u = - 2 x^{3}

= 5 (- \frac{1}{6} e^{- 2 x^{3}})

Simplify

5 (- \frac{1}{6} e^{- 2 x^{3}}) : - \frac{5}{6} e^{- 2 x^{3}}

= - \frac{5}{6} e^{- 2 x^{3}}

Add a constant to the solution

= - \frac{5}{6} e^{- 2 x^{3}} + C

\int 2 e^{x^{2}} x d x

\int (- x^{5} + 9) d x

\int cos^{2} (π x) sin (π x) d x

\int e^{t^{4}} d t

\int x e^{\frac{x}{a}} d x

What is the integral of (5x^2)/(e^{2x^3)} ?
The integral of (5x^2)/(e^{2x^3)} is -5/6 e^{-2x^3}+C