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

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

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

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

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

Solution steps

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

Apply u-substitution

= \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})

Substitute back

u = 1 - t

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

Simplify

= \frac{1}{1 - t}

Add a constant to the solution

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