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

The integral of x/((2x+1)^3) is (-4x-1)/(8(2x+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 x/((2x+1)^3)

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

\frac{- 4 x - 1}{8 ( 2 x + 1 ) ^{2}} + C

Solution steps

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

Apply u-substitution

: \int \frac{u - 1}{4 u ^{3}} d u

= \int \frac{u - 1}{4 u ^{3}} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{u - 1}{u ^{3}} d u

Expand

\frac{u - 1}{u ^{3}} : \frac{1}{u ^{2}} - \frac{1}{u ^{3}}

= \frac{1}{4} \cdot \int \frac{1}{u ^{2}} - \frac{1}{u ^{3}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{4} (\int \frac{1}{u ^{2}} d u - \int \frac{1}{u ^{3}} d u)

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

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

= \frac{1}{4} (- \frac{1}{u} - (- \frac{1}{2 u ^{2}}))

Substitute back

u = 2 x + 1

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

Simplify

\frac{1}{4} (- \frac{1}{2 x + 1} - (- \frac{1}{2 ( 2 x + 1 ) ^{2}})) : \frac{- 4 x - 1}{8 ( 2 x + 1 ) ^{2}}

= \frac{- 4 x - 1}{8 ( 2 x + 1 ) ^{2}}

Add a constant to the solution

= \frac{- 4 x - 1}{8 ( 2 x + 1 ) ^{2}} + C

\frac{d x}{x - x ^{2}} = 3 d t

\frac{\partial}{\partial y} (14 - x^{2} - 4 y^{2})

x \to 2 lim (\frac{x ^{2} + 1}{x})

\frac{d}{d x} (e^{4} + ln (8))

\int e^{3 x + 4} d x

What is the integral of x/((2x+1)^3) ?
The integral of x/((2x+1)^3) is (-4x-1)/(8(2x+1)^2)+C