What is the integral of 1/((x+6)(x+8)) ?

The integral of 1/((x+6)(x+8)) is 1/2 ln|x+6|-1/2 ln|x+8|+C

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integral of 1/((x+6)(x+8))

\int \frac{1}{( x + 6 ) ( x + 8 )} d x

\frac{1}{2} ln ∣ x + 6 ∣ - \frac{1}{2} ln ∣ x + 8 ∣ + C

Solution steps

\int \frac{1}{( x + 6 ) ( x + 8 )} d x

Take the partial fraction of

\frac{1}{( x + 6 ) ( x + 8 )} : \frac{1}{2 ( x + 6 )} - \frac{1}{2 ( x + 8 )}

= \int \frac{1}{2 ( x + 6 )} - \frac{1}{2 ( x + 8 )} 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

= \int \frac{1}{2 ( x + 6 )} d x - \int \frac{1}{2 ( x + 8 )} d x

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

\int \frac{1}{2 ( x + 8 )} d x = \frac{1}{2} ln ∣ x + 8 ∣

= \frac{1}{2} ln ∣ x + 6 ∣ - \frac{1}{2} ln ∣ x + 8 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ x + 6 ∣ - \frac{1}{2} ln ∣ x + 8 ∣ + C

\int_{0}^{5} \frac{1}{x ^{\frac{11}{12}}} d x

\frac{d}{d x} (6 - 5 cos (\frac{π}{15} x))

\int_{0}^{π} 18 sin^{4} (x) d x

x y^{^{''}} + y^{^{'}} = 28 x

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

What is the integral of 1/((x+6)(x+8)) ?
The integral of 1/((x+6)(x+8)) is 1/2 ln|x+6|-1/2 ln|x+8|+C