What is the integral of 1/((2x+3)^3) ?

The integral of 1/((2x+3)^3) is -1/(4(2x+3)^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/((2x+3)^3)

\int \frac{1}{( 2 x + 3 ) ^{3}} d x

- \frac{1}{4 ( 2 x + 3 ) ^{2}} + C

Solution steps

\int \frac{1}{( 2 x + 3 ) ^{3}} d x

Apply u-substitution

: \int \frac{1}{2 u ^{3}} d u

= \int \frac{1}{2 u ^{3}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

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

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

Substitute back

u = 2 x + 3

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

Simplify

\frac{1}{2} (- \frac{1}{2 ( 2 x + 3 ) ^{2}}) : - \frac{1}{4 ( 2 x + 3 ) ^{2}}

= - \frac{1}{4 ( 2 x + 3 ) ^{2}}

Add a constant to the solution

= - \frac{1}{4 ( 2 x + 3 ) ^{2}} + C

\int 8 x^{2} ln (x) d x

\frac{\partial}{\partial x} (2 x^{2} y + 9 x y^{3})

\frac{d}{d t} (sin (t) + cos (t))

\frac{d y}{d x} + 3 y = 5

\frac{\partial}{\partial x} (\frac{1}{y - 3 x})

What is the integral of 1/((2x+3)^3) ?
The integral of 1/((2x+3)^3) is -1/(4(2x+3)^2)+C