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

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

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

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

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

Solution steps

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

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}{( 1 - x ) ^{2}} d x

Apply u-substitution

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

= \frac{1}{4} \cdot \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

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

Apply the Power Rule

: - \frac{1}{u}

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

Substitute back

u = 1 - x

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

Simplify

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

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

Add a constant to the solution

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

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