What is the integral of 1/((x+1)(x+5)) ?

The integral of 1/((x+1)(x+5)) is 1/4 ln|x+1|-1/4 ln|x+5|+C

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integral of 1/((x+1)(x+5))

\int \frac{1}{( x + 1 ) ( x + 5 )} d x

\frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x + 5 ∣ + C

Solution steps

\int \frac{1}{( x + 1 ) ( x + 5 )} d x

Take the partial fraction of

\frac{1}{( x + 1 ) ( x + 5 )} : \frac{1}{4 ( x + 1 )} - \frac{1}{4 ( x + 5 )}

= \int \frac{1}{4 ( x + 1 )} - \frac{1}{4 ( x + 5 )} 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}{4 ( x + 1 )} d x - \int \frac{1}{4 ( x + 5 )} d x

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

\int \frac{1}{4 ( x + 5 )} d x = \frac{1}{4} ln ∣ x + 5 ∣

= \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x + 5 ∣

Add a constant to the solution

= \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{4} ln ∣ x + 5 ∣ + C

\int tan^{3} (9 x) sec (9 x) d x

y = cos (a^{9} + x^{9})

y = 6 x^{2} - 3 x, at 1, 5

(8.1) (4.1)

\frac{\partial}{\partial y} (x^{4} + 2 x^{2} y)

What is the integral of 1/((x+1)(x+5)) ?
The integral of 1/((x+1)(x+5)) is 1/4 ln|x+1|-1/4 ln|x+5|+C