What is the integral of 2/((x+4)(x+6)) ?

The integral of 2/((x+4)(x+6)) is ln|x+4|-ln|x+6|+C

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integral of 2/((x+4)(x+6))

\int \frac{2}{( x + 4 ) ( x + 6 )} d x

ln ∣ x + 4 ∣ - ln ∣ x + 6 ∣ + C

Solution steps

\int \frac{2}{( x + 4 ) ( x + 6 )} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{( x + 4 ) ( x + 6 )} d x

Take the partial fraction of

\frac{1}{( x + 4 ) ( x + 6 )} : \frac{1}{2 ( x + 4 )} - \frac{1}{2 ( x + 6 )}

= 2 \cdot \int \frac{1}{2 ( x + 4 )} - \frac{1}{2 ( x + 6 )} 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

= 2 (\int \frac{1}{2 ( x + 4 )} d x - \int \frac{1}{2 ( x + 6 )} d x)

\int \frac{1}{2 ( x + 4 )} d x = \frac{1}{2} ln ∣ x + 4 ∣

\int \frac{1}{2 ( x + 6 )} d x = \frac{1}{2} ln ∣ x + 6 ∣

= 2 (\frac{1}{2} ln ∣ x + 4 ∣ - \frac{1}{2} ln ∣ x + 6 ∣)

Simplify

2 (\frac{1}{2} ln ∣ x + 4 ∣ - \frac{1}{2} ln ∣ x + 6 ∣) : ln ∣ x + 4 ∣ - ln ∣ x + 6 ∣

= ln ∣ x + 4 ∣ - ln ∣ x + 6 ∣

Add a constant to the solution

= ln ∣ x + 4 ∣ - ln ∣ x + 6 ∣ + C

\int \frac{3}{4 x ^{2} - 1} d x

\int \frac{5}{x 4 x ^{4} - 9} d x

\int \frac{cos ( 3 x ) sin ( 3 x )}{3} d x

\frac{( x ^{2} + 2 x - 1 )}{( 3 - x )}

\int \frac{tan ^{2} ( 6 x )}{csc ( 6 x )} d x

What is the integral of 2/((x+4)(x+6)) ?
The integral of 2/((x+4)(x+6)) is ln|x+4|-ln|x+6|+C