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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- \frac{1}{x - 1} + C

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = x - 1

= - \frac{1}{x - 1}

Add a constant to the solution

= - \frac{1}{x - 1} + C

y^{^{''}} + 9 y = cos (2 t)

h \to 0 lim (\frac{( ( 4 + h ) - 4 )}{h})

\frac{d}{d x} (a x e^{4 x})

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

\int 12 x z cos (x z^{2}) d z

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