What is the integral of (x^2+2x+3)/((x-6)(x^2+4)) ?

The integral of (x^2+2x+3)/((x-6)(x^2+4)) is 51/40 ln|x-6|+1/40 (-11/2 ln|x^2+4|+7arctan(x/2))+C

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integral of (x^2+2x+3)/((x-6)(x^2+4))

\int \frac{x ^{2} + 2 x + 3}{( x - 6 ) ( x ^{2} + 4 )} d x

\frac{51}{40} ln ∣ x - 6 ∣ + \frac{1}{40} (- \frac{11}{2} ln x^{2} + 4 + 7 arctan (\frac{x}{2})) + C

Solution steps

\int \frac{x ^{2} + 2 x + 3}{( x - 6 ) ( x ^{2} + 4 )} d x

Take the partial fraction of

\frac{x ^{2} + 2 x + 3}{( x - 6 ) ( x ^{2} + 4 )} : \frac{51}{40 ( x - 6 )} + \frac{- 11 x + 14}{40 ( x ^{2} + 4 )}

= \int \frac{51}{40 ( x - 6 )} + \frac{- 11 x + 14}{40 ( x ^{2} + 4 )} 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{51}{40 ( x - 6 )} d x + \int \frac{- 11 x + 14}{40 ( x ^{2} + 4 )} d x

\int \frac{51}{40 ( x - 6 )} d x = \frac{51}{40} ln ∣ x - 6 ∣

\int \frac{- 11 x + 14}{40 ( x ^{2} + 4 )} d x = \frac{1}{40} (- \frac{11}{2} ln x^{2} + 4 + 7 arctan (\frac{x}{2}))

= \frac{51}{40} ln ∣ x - 6 ∣ + \frac{1}{40} (- \frac{11}{2} ln x^{2} + 4 + 7 arctan (\frac{x}{2}))

Add a constant to the solution

= \frac{51}{40} ln ∣ x - 6 ∣ + \frac{1}{40} (- \frac{11}{2} ln x^{2} + 4 + 7 arctan (\frac{x}{2})) + C

What is the integral of (x^2+2x+3)/((x-6)(x^2+4)) ?
The integral of (x^2+2x+3)/((x-6)(x^2+4)) is 51/40 ln|x-6|+1/40 (-11/2 ln|x^2+4|+7arctan(x/2))+C