What is the integral of (2t+1)e^{t^2+t} ?

The integral of (2t+1)e^{t^2+t} is e^{t^2+t}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2t+1)e^{t^2+t}

\int (2 t + 1) e^{t^{2} + t} d t

e^{t^{2} + t} + C

Solution steps

\int (2 t + 1) e^{t^{2} + t} d t

Apply u-substitution

: \int e^{u} d u

= \int e^{u} d u

Use the common integral:

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

= e^{u}

Substitute back

u = t^{2} + t

= e^{t^{2} + t}

Add a constant to the solution

= e^{t^{2} + t} + C

\frac{d}{d x} (ln (5 x^{2} - 3 x + 4))

\frac{ln ( x )}{x ^{4}}

\int \frac{x}{x - 36} d x

y = 4 x - x^{2}

x \cdot sin (x) \cdot d x + cos (y) \cdot d y = 0

What is the integral of (2t+1)e^{t^2+t} ?
The integral of (2t+1)e^{t^2+t} is e^{t^2+t}+C