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

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

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

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

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

Solution steps

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

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

Simplify

= \frac{1}{4 - x}

Add a constant to the solution

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

(y) d x - (x + y^{2} - x^{2}) d y = 0

\int \frac{54}{1 - cos ( 9 x )} d x

(2 x)^{^{'}}

(ln (cos (x)))^{^{'}}

x \to 3 lim (\frac{x ^{2} - 5}{x - 3})