What is the integral of ((x+4))/(x^3+3x^2-10x) ?

The integral of ((x+4))/(x^3+3x^2-10x) is -2/5 ln|x|+3/7 ln|x-2|-1/35 ln|x+5|+C

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integral of ((x+4))/(x^3+3x^2-10x)

\int \frac{( x + 4 )}{x ^{3} + 3 x ^{2} - 10 x} d x

- \frac{2}{5} ln ∣ x ∣ + \frac{3}{7} ln ∣ x - 2 ∣ - \frac{1}{35} ln ∣ x + 5 ∣ + C

Solution steps

\int \frac{x + 4}{x ^{3} + 3 x ^{2} - 10 x} d x

Take the partial fraction of

\frac{x + 4}{x ^{3} + 3 x ^{2} - 10 x} : - \frac{2}{5 x} + \frac{3}{7 ( x - 2 )} - \frac{1}{35 ( x + 5 )}

= \int - \frac{2}{5 x} + \frac{3}{7 ( x - 2 )} - \frac{1}{35 ( 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

= - \int \frac{2}{5 x} d x + \int \frac{3}{7 ( x - 2 )} d x - \int \frac{1}{35 ( x + 5 )} d x

\int \frac{2}{5 x} d x = \frac{2}{5} ln ∣ x ∣

\int \frac{3}{7 ( x - 2 )} d x = \frac{3}{7} ln ∣ x - 2 ∣

\int \frac{1}{35 ( x + 5 )} d x = \frac{1}{35} ln ∣ x + 5 ∣

= - \frac{2}{5} ln ∣ x ∣ + \frac{3}{7} ln ∣ x - 2 ∣ - \frac{1}{35} ln ∣ x + 5 ∣

Add a constant to the solution

= - \frac{2}{5} ln ∣ x ∣ + \frac{3}{7} ln ∣ x - 2 ∣ - \frac{1}{35} ln ∣ x + 5 ∣ + C

\frac{d}{d x} (- 6 x^{2})

y^{^{'}} = (y + 5) (x + 2), y (0) = 1

\frac{10}{( s ) ( s ^{2} + 8 s + 25 )}

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

\int \frac{( x ^{2} + x - 10 )}{( 2 x - 3 ) ( x ^{2} + 4 )} d x

What is the integral of ((x+4))/(x^3+3x^2-10x) ?
The integral of ((x+4))/(x^3+3x^2-10x) is -2/5 ln|x|+3/7 ln|x-2|-1/35 ln|x+5|+C