What is the integral of (2x^2+x+19)/((x+1)(x^2+9)) ?

The integral of (2x^2+x+19)/((x+1)(x^2+9)) is 2ln|x+1|+1/3 arctan(x/3)+C

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integral of (2x^2+x+19)/((x+1)(x^2+9))

\int \frac{2 x ^{2} + x + 19}{( x + 1 ) ( x ^{2} + 9 )} d x

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

Solution steps

\int \frac{2 x ^{2} + x + 19}{( x + 1 ) ( x ^{2} + 9 )} d x

Take the partial fraction of

\frac{2 x ^{2} + x + 19}{( x + 1 ) ( x ^{2} + 9 )} : \frac{2}{x + 1} + \frac{1}{x ^{2} + 9}

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

\int \frac{2}{x + 1} d x = 2 ln ∣ x + 1 ∣

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

= 2 ln ∣ x + 1 ∣ + \frac{1}{3} arctan (\frac{x}{3})

Add a constant to the solution

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

f (x) = 3 x^{2}

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{x}}{1 + x ^{2}}

\int_{0}^{2} x + 2 d x

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

\frac{\partial}{\partial z} (x e^{6 y} sin (5 z))

What is the integral of (2x^2+x+19)/((x+1)(x^2+9)) ?
The integral of (2x^2+x+19)/((x+1)(x^2+9)) is 2ln|x+1|+1/3 arctan(x/3)+C