What is the integral of (5-x^2)/((x^2+9)(3-x)) ?

The integral of (5-x^2)/((x^2+9)(3-x)) is 1/9 (7/2 ln|x^2+9|+7arctan(x/3))+2/9 ln|x-3|+C

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integral of (5-x^2)/((x^2+9)(3-x))

\int \frac{5 - x ^{2}}{( x ^{2} + 9 ) ( 3 - x )} d x

\frac{1}{9} (\frac{7}{2} ln x^{2} + 9 + 7 arctan (\frac{x}{3})) + \frac{2}{9} ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{5 - x ^{2}}{( x ^{2} + 9 ) ( 3 - x )} d x

Take the partial fraction of

\frac{5 - x ^{2}}{( x ^{2} + 9 ) ( 3 - x )} : \frac{7 x + 21}{9 ( x ^{2} + 9 )} + \frac{2}{9 ( x - 3 )}

= \int \frac{7 x + 21}{9 ( x ^{2} + 9 )} + \frac{2}{9 ( x - 3 )} 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{7 x + 21}{9 ( x ^{2} + 9 )} d x + \int \frac{2}{9 ( x - 3 )} d x

\int \frac{7 x + 21}{9 ( x ^{2} + 9 )} d x = \frac{1}{9} (\frac{7}{2} ln x^{2} + 9 + 7 arctan (\frac{x}{3}))

\int \frac{2}{9 ( x - 3 )} d x = \frac{2}{9} ln ∣ x - 3 ∣

= \frac{1}{9} (\frac{7}{2} ln x^{2} + 9 + 7 arctan (\frac{x}{3})) + \frac{2}{9} ln ∣ x - 3 ∣

Add a constant to the solution

= \frac{1}{9} (\frac{7}{2} ln x^{2} + 9 + 7 arctan (\frac{x}{3})) + \frac{2}{9} ln ∣ x - 3 ∣ + C

\frac{2}{( s + 1 ) ( s + 2 ) ^{2}}

\frac{d y}{d x} = - \frac{e ^{x} y ^{2}}{10}

\int 3 ln^{2} (x) d x

\frac{d y}{d t} (t) - 2 y = 3 e^{t}

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

What is the integral of (5-x^2)/((x^2+9)(3-x)) ?
The integral of (5-x^2)/((x^2+9)(3-x)) is 1/9 (7/2 ln|x^2+9|+7arctan(x/3))+2/9 ln|x-3|+C