What is the integral of (25)/(x^3-5x^2) ?

The integral of (25)/(x^3-5x^2) is -ln|x|+5/x+ln|x-5|+C

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integral of (25)/(x^3-5x^2)

\int \frac{25}{x ^{3} - 5 x ^{2}} d x

- ln ∣ x ∣ + \frac{5}{x} + ln ∣ x - 5 ∣ + C

Solution steps

\int \frac{25}{x ^{3} - 5 x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 25 \cdot \int \frac{1}{x ^{3} - 5 x ^{2}} d x

Take the partial fraction of

\frac{1}{x ^{3} - 5 x ^{2}} : - \frac{1}{25 x} - \frac{1}{5 x ^{2}} + \frac{1}{25 ( x - 5 )}

= 25 \cdot \int - \frac{1}{25 x} - \frac{1}{5 x ^{2}} + \frac{1}{25 ( x - 5 )} 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

= 25 (- \int \frac{1}{25 x} d x - \int \frac{1}{5 x ^{2}} d x + \int \frac{1}{25 ( x - 5 )} d x)

\int \frac{1}{25 x} d x = \frac{1}{25} ln ∣ x ∣

\int \frac{1}{5 x ^{2}} d x = - \frac{1}{5 x}

\int \frac{1}{25 ( x - 5 )} d x = \frac{1}{25} ln ∣ x - 5 ∣

= 25 (- \frac{1}{25} ln ∣ x ∣ - (- \frac{1}{5 x}) + \frac{1}{25} ln ∣ x - 5 ∣)

Simplify

25 (- \frac{1}{25} ln ∣ x ∣ - (- \frac{1}{5 x}) + \frac{1}{25} ln ∣ x - 5 ∣) : - ln ∣ x ∣ + \frac{5}{x} + ln ∣ x - 5 ∣

= - ln ∣ x ∣ + \frac{5}{x} + ln ∣ x - 5 ∣

Add a constant to the solution

= - ln ∣ x ∣ + \frac{5}{x} + ln ∣ x - 5 ∣ + C

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\frac{( s + 2 ) ^{2}}{( s + 1 ) ^{3}}

What is the integral of (25)/(x^3-5x^2) ?
The integral of (25)/(x^3-5x^2) is -ln|x|+5/x+ln|x-5|+C