What is the integral of (8x^2+x+73)/((x+1)(x^2+9)) ?

The integral of (8x^2+x+73)/((x+1)(x^2+9)) is 8ln|x+1|+1/3 arctan(x/3)+C

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integral of (8x^2+x+73)/((x+1)(x^2+9))

\int \frac{8 x ^{2} + x + 73}{( x + 1 ) ( x ^{2} + 9 )} d x

8 ln ∣ x + 1 ∣ + \frac{1}{3} arctan (\frac{x}{3}) + C

Solution steps

\int \frac{8 x ^{2} + x + 73}{( x + 1 ) ( x ^{2} + 9 )} d x

Take the partial fraction of

\frac{8 x ^{2} + x + 73}{( x + 1 ) ( x ^{2} + 9 )} : \frac{8}{x + 1} + \frac{1}{x ^{2} + 9}

= \int \frac{8}{x + 1} + \frac{1}{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{8}{x + 1} d x + \int \frac{1}{x ^{2} + 9} d x

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

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

= 8 ln ∣ x + 1 ∣ + \frac{1}{3} arctan (\frac{x}{3})

Add a constant to the solution

= 8 ln ∣ x + 1 ∣ + \frac{1}{3} arctan (\frac{x}{3}) + C

7 (x^{3} - x)^{4}

\frac{d}{d x} (\frac{3 x - 8}{x ^{2} - 1})

y^{^{'}} + \frac{3 y}{x} = \frac{3}{x}

y = \frac{9}{x ^{- 6}}

{\frac{s}{( s - 2 ) ( s - 3 ) ( s - 6 )}}

What is the integral of (8x^2+x+73)/((x+1)(x^2+9)) ?
The integral of (8x^2+x+73)/((x+1)(x^2+9)) is 8ln|x+1|+1/3 arctan(x/3)+C