What is the integral of 1/((3x+1)^6) ?

The integral of 1/((3x+1)^6) is -1/(15(3x+1)^5)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/((3x+1)^6)

\int \frac{1}{( 3 x + 1 ) ^{6}} d x

- \frac{1}{15 ( 3 x + 1 ) ^{5}} + C

Solution steps

\int \frac{1}{( 3 x + 1 ) ^{6}} d x

Apply u-substitution

: \int \frac{1}{3 u ^{6}} d u

= \int \frac{1}{3 u ^{6}} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u ^{6}} d u

Apply the Power Rule

: - \frac{1}{5 u ^{5}}

= \frac{1}{3} (- \frac{1}{5 u ^{5}})

Substitute back

u = 3 x + 1

= \frac{1}{3} (- \frac{1}{5 ( 3 x + 1 ) ^{5}})

Simplify

\frac{1}{3} (- \frac{1}{5 ( 3 x + 1 ) ^{5}}) : - \frac{1}{15 ( 3 x + 1 ) ^{5}}

= - \frac{1}{15 ( 3 x + 1 ) ^{5}}

Add a constant to the solution

= - \frac{1}{15 ( 3 x + 1 ) ^{5}} + C

\int cos (13 - 2 y) d y

\int \frac{x ^{3}}{( x + 1 ) ^{4}} d x

\int sin^{5} (x) \cdot cos^{6} (x) d x

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

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

What is the integral of 1/((3x+1)^6) ?
The integral of 1/((3x+1)^6) is -1/(15(3x+1)^5)+C