What is the integral of 1/(5x^3-7x^2) ?

The integral of 1/(5x^3-7x^2) is -5/49 ln|x|+1/(7x)+5/49 ln|5x-7|+C

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integral of 1/(5x^3-7x^2)

\int \frac{1}{5 x ^{3} - 7 x ^{2}} d x

- \frac{5}{49} ln ∣ x ∣ + \frac{1}{7 x} + \frac{5}{49} ln ∣ 5 x - 7 ∣ + C

Solution steps

\int \frac{1}{5 x ^{3} - 7 x ^{2}} d x

Take the partial fraction of

\frac{1}{5 x ^{3} - 7 x ^{2}} : - \frac{5}{49 x} - \frac{1}{7 x ^{2}} + \frac{25}{49 ( 5 x - 7 )}

= \int - \frac{5}{49 x} - \frac{1}{7 x ^{2}} + \frac{25}{49 ( 5 x - 7 )} 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{5}{49 x} d x - \int \frac{1}{7 x ^{2}} d x + \int \frac{25}{49 ( 5 x - 7 )} d x

\int \frac{5}{49 x} d x = \frac{5}{49} ln ∣ x ∣

\int \frac{1}{7 x ^{2}} d x = - \frac{1}{7 x}

\int \frac{25}{49 ( 5 x - 7 )} d x = \frac{5}{49} ln ∣ 5 x - 7 ∣

= - \frac{5}{49} ln ∣ x ∣ - (- \frac{1}{7 x}) + \frac{5}{49} ln ∣ 5 x - 7 ∣

Simplify

= - \frac{5}{49} ln ∣ x ∣ + \frac{1}{7 x} + \frac{5}{49} ln ∣ 5 x - 7 ∣

Add a constant to the solution

= - \frac{5}{49} ln ∣ x ∣ + \frac{1}{7 x} + \frac{5}{49} ln ∣ 5 x - 7 ∣ + C

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What is the integral of 1/(5x^3-7x^2) ?
The integral of 1/(5x^3-7x^2) is -5/49 ln|x|+1/(7x)+5/49 ln|5x-7|+C