What is the integral of e^{8x-3} ?

The integral of e^{8x-3} is 1/8 e^{8x-3}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^{8x-3}

\int e^{8 x - 3} d x

\frac{1}{8} e^{8 x - 3} + C

Solution steps

\int e^{8 x - 3} d x

Apply u-substitution

: \int e^{u} \frac{1}{8} d u

= \int e^{u} \frac{1}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{8} e^{u}

Substitute back

u = 8 x - 3

= \frac{1}{8} e^{8 x - 3}

Add a constant to the solution

= \frac{1}{8} e^{8 x - 3} + C

\int (2 x + 1) 2 d x

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

\int \frac{1}{1 - 36 t ^{2}} d t

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

\int (48 x^{3} + 12) (2 x^{4} + 2 x)^{29} d x