What is the integral of 1/(x^2+10x+24) ?

The integral of 1/(x^2+10x+24) is 1/2 ln|x+4|-1/2 ln|x+6|+C

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integral of 1/(x^2+10x+24)

\int \frac{1}{x ^{2} + 10 x + 24} d x

\frac{1}{2} ln ∣ x + 4 ∣ - \frac{1}{2} ln ∣ x + 6 ∣ + C

Solution steps

\int \frac{1}{x ^{2} + 10 x + 24} d x

Take the partial fraction of

\frac{1}{x ^{2} + 10 x + 24} : \frac{1}{2 ( x + 4 )} - \frac{1}{2 ( x + 6 )}

= \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

= \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 ∣

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

Add a constant to the solution

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

\int \frac{1}{x ^{\frac{1}{5}}} d x

\frac{d}{d x} (\frac{( sin ( x ) )}{x})

\frac{d}{d x} (9 tan (7 x))

\frac{4 e ^{(- s)}}{( s ^{2} + 4 s + 13 )}

\int \frac{1}{x ^{2} - 6 x - 13} d x

What is the integral of 1/(x^2+10x+24) ?
The integral of 1/(x^2+10x+24) is 1/2 ln|x+4|-1/2 ln|x+6|+C