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

The integral of 1/((3x+1)^3) is -1/(6(3x+1)^2)+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)^3)

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

- \frac{1}{6 ( 3 x + 1 ) ^{2}} + C

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= \frac{1}{3} (- \frac{1}{2 u ^{2}})

Substitute back

u = 3 x + 1

= \frac{1}{3} (- \frac{1}{2 ( 3 x + 1 ) ^{2}})

Simplify

\frac{1}{3} (- \frac{1}{2 ( 3 x + 1 ) ^{2}}) : - \frac{1}{6 ( 3 x + 1 ) ^{2}}

= - \frac{1}{6 ( 3 x + 1 ) ^{2}}

Add a constant to the solution

= - \frac{1}{6 ( 3 x + 1 ) ^{2}} + C

\int (\frac{6}{1 - x ^{2}}) d x

\int \frac{3 x ^{4} - 5 x ^{3} + 7 x - 1}{x ^{\frac{3}{4}}} d x

\int (x^{2} - 6 x + 13)^{- 1} d x

\int x^{2} (x^{2} - 16)^{\frac{3}{2}} d x

\int + cos (9 x) sin (5 x) d x

What is the integral of 1/((3x+1)^3) ?
The integral of 1/((3x+1)^3) is -1/(6(3x+1)^2)+C