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

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

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

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

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

Solution steps

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

Apply u-substitution

= 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

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

Substitute back

u = 4 - x

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

Simplify

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

= \frac{4}{4 - x}

Add a constant to the solution

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

\int \frac{8 x}{x ^{5}} d x

\int 9000 e^{0.06 x} d x

\int \frac{1}{tanh ( x )} d x

\int cos (\frac{π}{2}) t d t

\int \frac{x ^{2}}{3 x} d x