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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/((4x-1)^2)

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

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

Solution steps

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

Apply u-substitution

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

Substitute back

u = 4 x - 1

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

Simplify

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

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

Add a constant to the solution

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

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

\int \frac{1 + e ^{- x}}{e ^{- x}} d x

\int 10 + 6 x y d x

\int 0.05 x d x

\int \frac{( x ^{- \frac{1}{3}} - 5 )}{x} d x

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