What is the integral of ye^{3y^2-1} ?

The integral of ye^{3y^2-1} is 1/6 e^{3y^2-1}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ye^{3y^2-1}

\int y e^{3 y^{2} - 1} d y

\frac{1}{6} e^{3 y^{2} - 1} + C

Solution steps

\int y e^{3 y^{2} - 1} d y

Apply u-substitution

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

= \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

= \frac{1}{6} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{6} e^{u}

Substitute back

u = 3 y^{2} - 1

= \frac{1}{6} e^{3 y^{2} - 1}

Add a constant to the solution

= \frac{1}{6} e^{3 y^{2} - 1} + C

\int \frac{1}{4 e ^{- 2 x} - 25} d x

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

\int 12.3 d x

\int (3 x + 1) cos (3 x + 2) d x

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

What is the integral of ye^{3y^2-1} ?
The integral of ye^{3y^2-1} is 1/6 e^{3y^2-1}+C