What is the integral of ((3x-5))/(x^3-5x^2-14x) ?

The integral of ((3x-5))/(x^3-5x^2-14x) is 5/14 ln|x|-11/18 ln|x+2|+16/63 ln|x-7|+C

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integral of ((3x-5))/(x^3-5x^2-14x)

\int \frac{( 3 x - 5 )}{x ^{3} - 5 x ^{2} - 14 x} d x

\frac{5}{14} ln ∣ x ∣ - \frac{11}{18} ln ∣ x + 2 ∣ + \frac{16}{63} ln ∣ x - 7 ∣ + C

Solution steps

\int \frac{3 x - 5}{x ^{3} - 5 x ^{2} - 14 x} d x

Take the partial fraction of

\frac{3 x - 5}{x ^{3} - 5 x ^{2} - 14 x} : \frac{5}{14 x} - \frac{11}{18 ( x + 2 )} + \frac{16}{63 ( x - 7 )}

= \int \frac{5}{14 x} - \frac{11}{18 ( x + 2 )} + \frac{16}{63 ( 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}{14 x} d x - \int \frac{11}{18 ( x + 2 )} d x + \int \frac{16}{63 ( x - 7 )} d x

\int \frac{5}{14 x} d x = \frac{5}{14} ln ∣ x ∣

\int \frac{11}{18 ( x + 2 )} d x = \frac{11}{18} ln ∣ x + 2 ∣

\int \frac{16}{63 ( x - 7 )} d x = \frac{16}{63} ln ∣ x - 7 ∣

= \frac{5}{14} ln ∣ x ∣ - \frac{11}{18} ln ∣ x + 2 ∣ + \frac{16}{63} ln ∣ x - 7 ∣

Add a constant to the solution

= \frac{5}{14} ln ∣ x ∣ - \frac{11}{18} ln ∣ x + 2 ∣ + \frac{16}{63} ln ∣ x - 7 ∣ + C

\frac{d}{d x} (3 x^{3} + y^{3})

(x cos (x))^{^{'}}

\int cos (x) sec^{2} (x) d x

f (x) = - sec^{2} (x)

\int_{0}^{\frac{π}{4}} 4 sec^{3} (x) d x

What is the integral of ((3x-5))/(x^3-5x^2-14x) ?
The integral of ((3x-5))/(x^3-5x^2-14x) is 5/14 ln|x|-11/18 ln|x+2|+16/63 ln|x-7|+C