What is the integral of 4/((3-x)^2) ?

The integral of 4/((3-x)^2) is 4/(3-x)+C

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integral of 4/((3-x)^2)

\int \frac{4}{( 3 - x ) ^{2}} d x

\frac{4}{3 - x} + C

Solution steps

\int \frac{4}{( 3 - x ) ^{2}} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{( 3 - x ) ^{2}} d x

Apply u-substitution

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

= 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

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

Apply the Power Rule

: - \frac{1}{u}

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

Substitute back

u = 3 - x

= 4 (- (- \frac{1}{3 - x}))

Simplify

4 (- (- \frac{1}{3 - x})) : \frac{4}{3 - x}

= \frac{4}{3 - x}

Add a constant to the solution

= \frac{4}{3 - x} + C

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