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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int \frac{1}{( 4 x + 3 ) ^{2}} d x

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

Solution steps

\int \frac{1}{( 4 x + 3 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{4 u ^{2}} d u

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

Apply the Power Rule

: - \frac{1}{u}

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

Substitute back

u = 4 x + 3

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

Simplify

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

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

Add a constant to the solution

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

\int - x sin (z) d z

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

\frac{d}{d x} (\frac{x + 5}{x - 3})

\frac{d}{d y} (y^{\frac{4}{5}})

y^{^{'}} = y (x y^{4} + 5)

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