What is the integral of te^{6t^2} ?

The integral of te^{6t^2} is 1/12 e^{6t^2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of te^{6t^2}

\int t e^{6 t^{2}} d t

\frac{1}{12} e^{6 t^{2}} + C

Solution steps

\int t e^{6 t^{2}} d t

Apply u-substitution

= \int \frac{e ^{u}}{12} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{12} e^{u}

Substitute back

u = 6 t^{2}

= \frac{1}{12} e^{6 t^{2}}

Add a constant to the solution

= \frac{1}{12} e^{6 t^{2}} + C

What is the integral of te^{6t^2} ?
The integral of te^{6t^2} is 1/12 e^{6t^2}+C