What is the integral of 1/((x-1)(x-3)(x+4)) ?

The integral of 1/((x-1)(x-3)(x+4)) is -1/10 ln|x-1|+1/14 ln|x-3|+1/35 ln|x+4|+C

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integral of 1/((x-1)(x-3)(x+4))

\int \frac{1}{( x - 1 ) ( x - 3 ) ( x + 4 )} d x

- \frac{1}{10} ln ∣ x - 1 ∣ + \frac{1}{14} ln ∣ x - 3 ∣ + \frac{1}{35} ln ∣ x + 4 ∣ + C

Solution steps

\int \frac{1}{( x - 1 ) ( x - 3 ) ( x + 4 )} d x

Take the partial fraction of

\frac{1}{( x - 1 ) ( x - 3 ) ( x + 4 )} : - \frac{1}{10 ( x - 1 )} + \frac{1}{14 ( x - 3 )} + \frac{1}{35 ( x + 4 )}

= \int - \frac{1}{10 ( x - 1 )} + \frac{1}{14 ( x - 3 )} + \frac{1}{35 ( x + 4 )} 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{1}{10 ( x - 1 )} d x + \int \frac{1}{14 ( x - 3 )} d x + \int \frac{1}{35 ( x + 4 )} d x

\int \frac{1}{10 ( x - 1 )} d x = \frac{1}{10} ln ∣ x - 1 ∣

\int \frac{1}{14 ( x - 3 )} d x = \frac{1}{14} ln ∣ x - 3 ∣

\int \frac{1}{35 ( x + 4 )} d x = \frac{1}{35} ln ∣ x + 4 ∣

= - \frac{1}{10} ln ∣ x - 1 ∣ + \frac{1}{14} ln ∣ x - 3 ∣ + \frac{1}{35} ln ∣ x + 4 ∣

Add a constant to the solution

= - \frac{1}{10} ln ∣ x - 1 ∣ + \frac{1}{14} ln ∣ x - 3 ∣ + \frac{1}{35} ln ∣ x + 4 ∣ + C

(9 - x^{2})^{^{'}}

f (x) = x e^{x^{2}}

\frac{d}{d x} (240 (1 - \frac{3}{3 + e ^{- 0.5 x}}))

e^{\frac{3}{2}}

\int x^{3} 3 + x^{4} d x

What is the integral of 1/((x-1)(x-3)(x+4)) ?
The integral of 1/((x-1)(x-3)(x+4)) is -1/10 ln|x-1|+1/14 ln|x-3|+1/35 ln|x+4|+C