What is the integral of (x^2-7)/(x^2(x+5)(x-3)) ?

The integral of (x^2-7)/(x^2(x+5)(x-3)) is 14/225 ln|x|-7/(15x)-9/100 ln|x+5|+1/36 ln|x-3|+C

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integral of (x^2-7)/(x^2(x+5)(x-3))

\int \frac{x ^{2} - 7}{x ^{2} ( x + 5 ) ( x - 3 )} d x

\frac{14}{225} ln ∣ x ∣ - \frac{7}{15 x} - \frac{9}{100} ln ∣ x + 5 ∣ + \frac{1}{36} ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{x ^{2} - 7}{x ^{2} ( x + 5 ) ( x - 3 )} d x

Take the partial fraction of

\frac{x ^{2} - 7}{x ^{2} ( x + 5 ) ( x - 3 )} : \frac{14}{225 x} + \frac{7}{15 x ^{2}} - \frac{9}{100 ( x + 5 )} + \frac{1}{36 ( x - 3 )}

= \int \frac{14}{225 x} + \frac{7}{15 x ^{2}} - \frac{9}{100 ( x + 5 )} + \frac{1}{36 ( 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{14}{225 x} d x + \int \frac{7}{15 x ^{2}} d x - \int \frac{9}{100 ( x + 5 )} d x + \int \frac{1}{36 ( x - 3 )} d x

\int \frac{14}{225 x} d x = \frac{14}{225} ln ∣ x ∣

\int \frac{7}{15 x ^{2}} d x = - \frac{7}{15 x}

\int \frac{9}{100 ( x + 5 )} d x = \frac{9}{100} ln ∣ x + 5 ∣

\int \frac{1}{36 ( x - 3 )} d x = \frac{1}{36} ln ∣ x - 3 ∣

= \frac{14}{225} ln ∣ x ∣ - \frac{7}{15 x} - \frac{9}{100} ln ∣ x + 5 ∣ + \frac{1}{36} ln ∣ x - 3 ∣

Add a constant to the solution

= \frac{14}{225} ln ∣ x ∣ - \frac{7}{15 x} - \frac{9}{100} ln ∣ x + 5 ∣ + \frac{1}{36} ln ∣ x - 3 ∣ + C

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What is the integral of (x^2-7)/(x^2(x+5)(x-3)) ?
The integral of (x^2-7)/(x^2(x+5)(x-3)) is 14/225 ln|x|-7/(15x)-9/100 ln|x+5|+1/36 ln|x-3|+C