What is the integral of (3x-17)/((x^2+9)(x+1)) ?

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

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integral of (3x-17)/((x^2+9)(x+1))

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

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

\int \frac{1}{x ^{2} 9 x ^{2} + 64} d x

\int x e^{3 - x^{2}} d x

\int \frac{3 x}{4 x ^{2} + 5} d x

\int \frac{13 ( tan ( x ) ) ^{2}}{sec ( x )} d x

e^{y^{2}}

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