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

\frac{d}{d x} (ln (arctan (x)))

\frac{d y}{d x} - 3 y = 0

n = 1 \sum \infty (- 0.7)^{n - 1}

\frac{\partial}{\partial t} (\frac{x}{t})

\int_{0}^{3} π (9 x^{2} - x^{4}) 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