What is the integral of (6x^2+x+55)/((x+1)(x^2+9)) ?

The integral of (6x^2+x+55)/((x+1)(x^2+9)) is 6ln|x+1|+1/3 arctan(x/3)+C

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integral of (6x^2+x+55)/((x+1)(x^2+9))

\int \frac{6 x ^{2} + x + 55}{( x + 1 ) ( x ^{2} + 9 )} d x

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

Solution steps

\int \frac{6 x ^{2} + x + 55}{( x + 1 ) ( x ^{2} + 9 )} d x

Take the partial fraction of

\frac{6 x ^{2} + x + 55}{( x + 1 ) ( x ^{2} + 9 )} : \frac{6}{x + 1} + \frac{1}{x ^{2} + 9}

= \int \frac{6}{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{6}{x + 1} d x + \int \frac{1}{x ^{2} + 9} d x

\int \frac{6}{x + 1} d x = 6 ln ∣ x + 1 ∣

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

= 6 ln ∣ x + 1 ∣ + \frac{1}{3} arctan (\frac{x}{3})

Add a constant to the solution

= 6 ln ∣ x + 1 ∣ + \frac{1}{3} arctan (\frac{x}{3}) + C

\int y (1 - x^{2}) d y

\frac{d}{d x} ((5 x - 1) (6 x - 3)^{- 1})

\frac{d}{d x} (1 + 3^{x})

x \to 2 π lim (x)

What is the integral of (6x^2+x+55)/((x+1)(x^2+9)) ?
The integral of (6x^2+x+55)/((x+1)(x^2+9)) is 6ln|x+1|+1/3 arctan(x/3)+C