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

The integral of (2x^2+x+1)/((x+1)(x^2+9)) is 1/5 ln|x+1|+1/5 (9/2 ln|x^2+9|-4/3 arctan(x/3))+C

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

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

\frac{1}{5} ln ∣ x + 1 ∣ + \frac{1}{5} (\frac{9}{2} ln x^{2} + 9 - \frac{4}{3} arctan (\frac{x}{3})) + C

Solution steps

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

Take the partial fraction of

\frac{2 x ^{2} + x + 1}{( x + 1 ) ( x ^{2} + 9 )} : \frac{1}{5 ( x + 1 )} + \frac{9 x - 4}{5 ( x ^{2} + 9 )}

= \int \frac{1}{5 ( x + 1 )} + \frac{9 x - 4}{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{1}{5 ( x + 1 )} d x + \int \frac{9 x - 4}{5 ( x ^{2} + 9 )} d x

\int \frac{1}{5 ( x + 1 )} d x = \frac{1}{5} ln ∣ x + 1 ∣

\int \frac{9 x - 4}{5 ( x ^{2} + 9 )} d x = \frac{1}{5} (\frac{9}{2} ln x^{2} + 9 - \frac{4}{3} arctan (\frac{x}{3}))

= \frac{1}{5} ln ∣ x + 1 ∣ + \frac{1}{5} (\frac{9}{2} ln x^{2} + 9 - \frac{4}{3} arctan (\frac{x}{3}))

Add a constant to the solution

= \frac{1}{5} ln ∣ x + 1 ∣ + \frac{1}{5} (\frac{9}{2} ln x^{2} + 9 - \frac{4}{3} arctan (\frac{x}{3})) + C

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What is the integral of (2x^2+x+1)/((x+1)(x^2+9)) ?
The integral of (2x^2+x+1)/((x+1)(x^2+9)) is 1/5 ln|x+1|+1/5 (9/2 ln|x^2+9|-4/3 arctan(x/3))+C