What is the integral of 4/((x-7)(x-3)) ?

The integral of 4/((x-7)(x-3)) is ln|x-7|-ln|x-3|+C

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integral of 4/((x-7)(x-3))

\int \frac{4}{( x - 7 ) ( x - 3 )} d x

ln ∣ x - 7 ∣ - ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{4}{( x - 7 ) ( x - 3 )} 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}{( x - 7 ) ( x - 3 )} d x

Take the partial fraction of

\frac{1}{( x - 7 ) ( x - 3 )} : \frac{1}{4 ( x - 7 )} - \frac{1}{4 ( x - 3 )}

= 4 \cdot \int \frac{1}{4 ( x - 7 )} - \frac{1}{4 ( x - 3 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (\int \frac{1}{4 ( x - 7 )} d x - \int \frac{1}{4 ( x - 3 )} d x)

\int \frac{1}{4 ( x - 7 )} d x = \frac{1}{4} ln ∣ x - 7 ∣

\int \frac{1}{4 ( x - 3 )} d x = \frac{1}{4} ln ∣ x - 3 ∣

= 4 (\frac{1}{4} ln ∣ x - 7 ∣ - \frac{1}{4} ln ∣ x - 3 ∣)

Simplify

4 (\frac{1}{4} ln ∣ x - 7 ∣ - \frac{1}{4} ln ∣ x - 3 ∣) : ln ∣ x - 7 ∣ - ln ∣ x - 3 ∣

= ln ∣ x - 7 ∣ - ln ∣ x - 3 ∣

Add a constant to the solution

= ln ∣ x - 7 ∣ - ln ∣ x - 3 ∣ + C

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

x^{2} cos (x)

\int tan (2 x) sec^{3} (2 x) d x

\int \frac{( cos ( ln ( x ) ) )}{x} d x

\int tan (3 x + 1) d x

What is the integral of 4/((x-7)(x-3)) ?
The integral of 4/((x-7)(x-3)) is ln|x-7|-ln|x-3|+C