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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{u}

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

Substitute back

u = 1 - x

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

Simplify

= \frac{1}{1 - x}

Add a constant to the solution

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

\int 2 t + 5 d t

\int \frac{2 x}{x ^{2} + y ^{2}} d y

\frac{d}{d x} (7 x^{3} + 4)

\int \frac{18}{x ^{2} + 4 x - 5} d x

\frac{1}{( s ^{3} + 1 ) ( s ^{2} + 1 )}

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