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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Apply u-substitution

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

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

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

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

\int 3 d u = 3 u

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

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

Substitute back

u = 2 x + 3

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

Simplify

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

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

Add a constant to the solution

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

\int x^{2} e^{x^{3} + 8} d x

\int \frac{1}{2} x^{2} \cdot e^{- x} d x

\int x (10 - x)^{\frac{1}{2}} d x

\int (x + 3)^{9} d x

\int x^{3} a^{2} x^{2} - b^{2} d x

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