What is the integral of 6^{x+1} ?

The integral of 6^{x+1} is (6^{x+1})/(ln(6))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 6^{x+1}

\int 6^{x + 1} d x

\frac{6 ^{x + 1}}{ln ( 6 )} + C

Solution steps

\int 6^{x + 1} d x

Simplify

6^{x + 1} : 6^{x} \cdot 6

= \int 6^{x} \cdot 6 d x

Take the constant out:

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

= 6 \cdot \int 6^{x} d x

\int a^{x} d x = \frac{a ^{x}}{ln a}

= 6 \cdot \frac{6 ^{x}}{ln ( 6 )}

Simplify

6 \cdot \frac{6 ^{x}}{ln ( 6 )} : \frac{6 ^{x + 1}}{ln ( 6 )}

= \frac{6 ^{x + 1}}{ln ( 6 )}

Add a constant to the solution

= \frac{6 ^{x + 1}}{ln ( 6 )} + C

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

\int (1 - 3 x)^{7} d x

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

\int 2 x e^{2 y} - x cos (x y) + 2 y d y

\int - \frac{1}{1 + e ^{- x}} d x

What is the integral of 6^{x+1} ?
The integral of 6^{x+1} is (6^{x+1})/(ln(6))+C