What is the integral of 1/(12+4x-x^2) ?

The integral of 1/(12+4x-x^2) is 1/8 ln|x+2|-1/8 ln|x-6|+C

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integral of 1/(12+4x-x^2)

\int \frac{1}{12 + 4 x - x ^{2}} d x

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

Solution steps

\int \frac{1}{12 + 4 x - x ^{2}} d x

Take the partial fraction of

\frac{1}{12 + 4 x - x ^{2}} : \frac{1}{8 ( x + 2 )} - \frac{1}{8 ( x - 6 )}

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

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

\int \frac{1}{8 ( x - 6 )} d x = \frac{1}{8} ln ∣ x - 6 ∣

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

Add a constant to the solution

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

\frac{1}{( x ^{2} + 4 ) ^{2}}

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What is the integral of 1/(12+4x-x^2) ?
The integral of 1/(12+4x-x^2) is 1/8 ln|x+2|-1/8 ln|x-6|+C