What is the integral of (6x^2+x+16)/((x^2+3)(x-2)) ?

The integral of (6x^2+x+16)/((x^2+3)(x-2)) is

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integral of (6x^2+x+16)/((x^2+3)(x-2))

\int \frac{6 x ^{2} + x + 16}{( x ^{2} + 3 ) ( x - 2 )} d x

3 ln \frac{x ^{2}}{3} + 1 - 43 arctan (\frac{x}{3}) + \frac{25}{3} arctan (\frac{1}{3} x) + 42 (\frac{1}{7} ln ∣ x - 2 ∣ - \frac{1}{14} ln \frac{1}{3} x^{2} + 1 - \frac{2}{7 3} arctan (\frac{1}{3} x)) + C

Solution steps

\int \frac{6 x ^{2} + x + 16}{( x ^{2} + 3 ) ( x - 2 )} d x

Apply u-substitution

: \int \frac{6 u ^{2} + 25 u + 42}{u ( u ^{2} + 4 u + 7 )} d u

= \int \frac{6 u ^{2} + 25 u + 42}{u ( u ^{2} + 4 u + 7 )} d u

Expand

\frac{6 u ^{2} + 25 u + 42}{u ( u ^{2} + 4 u + 7 )} : \frac{6 u ^{2}}{u ( u ^{2} + 4 u + 7 )} + \frac{25 u}{u ( u ^{2} + 4 u + 7 )} + \frac{42}{u ( u ^{2} + 4 u + 7 )}

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{6 u ^{2}}{u ( u ^{2} + 4 u + 7 )} d u + \int \frac{25 u}{u ( u ^{2} + 4 u + 7 )} d u + \int \frac{42}{u ( u ^{2} + 4 u + 7 )} d u

\int \frac{6 u ^{2}}{u ( u ^{2} + 4 u + 7 )} d u = 3 ln \frac{1}{3} (u + 2)^{2} + 1 - 43 arctan (\frac{1}{3} (u + 2))

\int \frac{25 u}{u ( u ^{2} + 4 u + 7 )} d u = \frac{25}{3} arctan (\frac{1}{3} (u + 2))

\int \frac{42}{u ( u ^{2} + 4 u + 7 )} d u = 42 (\frac{1}{7} ln ∣ u ∣ - \frac{1}{14} ln \frac{1}{3} (u + 2)^{2} + 1 - \frac{2}{7 3} arctan (\frac{1}{3} (u + 2)))

= 3 ln \frac{1}{3} (u + 2)^{2} + 1 - 43 arctan (\frac{1}{3} (u + 2)) + \frac{25}{3} arctan (\frac{1}{3} (u + 2)) + 42 (\frac{1}{7} ln ∣ u ∣ - \frac{1}{14} ln \frac{1}{3} (u + 2)^{2} + 1 - \frac{2}{7 3} arctan (\frac{1}{3} (u + 2)))

Substitute back

u = x - 2

= 3 ln \frac{1}{3} (x - 2 + 2)^{2} + 1 - 43 arctan (\frac{1}{3} (x - 2 + 2)) + \frac{25}{3} arctan (\frac{1}{3} (x - 2 + 2)) + 42 (\frac{1}{7} ln ∣ x - 2 ∣ - \frac{1}{14} ln \frac{1}{3} (x - 2 + 2)^{2} + 1 - \frac{2}{7 3} arctan (\frac{1}{3} (x - 2 + 2)))

Simplify

3 ln \frac{1}{3} (x - 2 + 2)^{2} + 1 - 43 arctan (\frac{1}{3} (x - 2 + 2)) + \frac{25}{3} arctan (\frac{1}{3} (x - 2 + 2)) + 42 (\frac{1}{7} ln ∣ x - 2 ∣ - \frac{1}{14} ln \frac{1}{3} (x - 2 + 2)^{2} + 1 - \frac{2}{7 3} arctan (\frac{1}{3} (x - 2 + 2))) : 3 ln \frac{x ^{2}}{3} + 1 - 43 arctan (\frac{x}{3}) + \frac{25}{3} arctan (\frac{1}{3} x) + 42 (\frac{1}{7} ln ∣ x - 2 ∣ - \frac{1}{14} ln \frac{1}{3} x^{2} + 1 - \frac{2}{7 3} arctan (\frac{1}{3} x))

= 3 ln \frac{x ^{2}}{3} + 1 - 43 arctan (\frac{x}{3}) + \frac{25}{3} arctan (\frac{1}{3} x) + 42 (\frac{1}{7} ln ∣ x - 2 ∣ - \frac{1}{14} ln \frac{1}{3} x^{2} + 1 - \frac{2}{7 3} arctan (\frac{1}{3} x))

Add a constant to the solution

= 3 ln \frac{x ^{2}}{3} + 1 - 43 arctan (\frac{x}{3}) + \frac{25}{3} arctan (\frac{1}{3} x) + 42 (\frac{1}{7} ln ∣ x - 2 ∣ - \frac{1}{14} ln \frac{1}{3} x^{2} + 1 - \frac{2}{7 3} arctan (\frac{1}{3} x)) + C

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What is the integral of (6x^2+x+16)/((x^2+3)(x-2)) ?
The integral of (6x^2+x+16)/((x^2+3)(x-2)) is