What is the integral of (5x^2-x+17)/((x+2)(x^2+9)) ?

The integral of (5x^2-x+17)/((x+2)(x^2+9)) is 3ln|x+2|+ln|(x^2)/9+1|-5/3 arctan(x/3)+C

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integral of (5x^2-x+17)/((x+2)(x^2+9))

\int \frac{5 x ^{2} - x + 17}{( x + 2 ) ( x ^{2} + 9 )} d x

3 ln ∣ x + 2 ∣ + ln \frac{x ^{2}}{9} + 1 - \frac{5}{3} arctan (\frac{x}{3}) + C

Solution steps

\int \frac{5 x ^{2} - x + 17}{( x + 2 ) ( x ^{2} + 9 )} d x

Take the partial fraction of

\frac{5 x ^{2} - x + 17}{( x + 2 ) ( x ^{2} + 9 )} : \frac{3}{x + 2} + \frac{2 x - 5}{x ^{2} + 9}

= \int \frac{3}{x + 2} + \frac{2 x - 5}{x ^{2} + 9} 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{3}{x + 2} d x + \int \frac{2 x - 5}{x ^{2} + 9} d x

\int \frac{3}{x + 2} d x = 3 ln ∣ x + 2 ∣

\int \frac{2 x - 5}{x ^{2} + 9} d x = ln \frac{x ^{2}}{9} + 1 - \frac{5}{3} arctan (\frac{x}{3})

= 3 ln ∣ x + 2 ∣ + ln \frac{x ^{2}}{9} + 1 - \frac{5}{3} arctan (\frac{x}{3})

Add a constant to the solution

= 3 ln ∣ x + 2 ∣ + ln \frac{x ^{2}}{9} + 1 - \frac{5}{3} arctan (\frac{x}{3}) + C

\frac{d y}{d x} = e^{8 x + y}

d er i v a t i v e f (x) = csc (x + 5)

f (x) = \frac{1}{3 x + 1}

y^{^{'}} = t e^{- y}, y (0) = 1

\int_{0}^{1} cos (2 π x) d x

What is the integral of (5x^2-x+17)/((x+2)(x^2+9)) ?
The integral of (5x^2-x+17)/((x+2)(x^2+9)) is 3ln|x+2|+ln|(x^2)/9+1|-5/3 arctan(x/3)+C