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

t an g e n t f (x) = - 2 sin (x), at x = \frac{3}{4} π

\frac{d}{d x} (\frac{1}{arccos ( x ^{2} + 2 x + 4 )})

\int 0 e^{x} d x

\int 6 e^{- 5 x} d x

t an g e n t f (x) = 1 - x^{2}, at x = 2

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