What is the integral of (2x^3+x+1)/(x^2+9) ?

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

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integral of (2x^3+x+1)/(x^2+9)

\int \frac{2 x ^{3} + x + 1}{x ^{2} + 9} d x

x^{2} - \frac{17}{2} ln x^{2} + 9 + \frac{1}{3} arctan (\frac{x}{3}) + C

Solution steps

\int \frac{2 x ^{3} + x + 1}{x ^{2} + 9} d x

Long division

\frac{2 x ^{3} + x + 1}{x ^{2} + 9} : 2 x + \frac{- 17 x + 1}{x ^{2} + 9}

= \int 2 x + \frac{- 17 x + 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 2 x d x + \int \frac{- 17 x + 1}{x ^{2} + 9} d x

\int 2 x d x = x^{2}

\int \frac{- 17 x + 1}{x ^{2} + 9} d x = - \frac{17}{2} ln x^{2} + 9 + \frac{1}{3} arctan (\frac{x}{3})

= x^{2} - \frac{17}{2} ln x^{2} + 9 + \frac{1}{3} arctan (\frac{x}{3})

Add a constant to the solution

= x^{2} - \frac{17}{2} ln x^{2} + 9 + \frac{1}{3} arctan (\frac{x}{3}) + C

1 - 9 x

\int_{0}^{4} 0.125 x^{3} d x

e^{- x} arctan (2 e^{x})

\int (3 x^{2} + 4 x + 1) d x

n = 0 \sum \infty (- x^{2} - 8)^{n}

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