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

The integral of 1/((x-1)(x+4)) is 1/5 ln|x-1|-1/5 ln|x+4|+C

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

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

\frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{( x - 1 ) ( x + 4 )} : \frac{1}{5 ( x - 1 )} - \frac{1}{5 ( x + 4 )}

= \int \frac{1}{5 ( x - 1 )} - \frac{1}{5 ( 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}{5 ( x - 1 )} d x - \int \frac{1}{5 ( x + 4 )} d x

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

\int \frac{1}{5 ( x + 4 )} d x = \frac{1}{5} ln ∣ x + 4 ∣

= \frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣

Add a constant to the solution

= \frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣ + C

x^{2} \cdot arctan (x)

x \to 0 + lim (5 x^{s i n (x)})

cos (x) \frac{π}{2}

y^{^{'}} = a y + b

\int_{0}^{\frac{1}{2}} \frac{4}{1 + 4 x ^{2}} d x

What is the integral of 1/((x-1)(x+4)) ?
The integral of 1/((x-1)(x+4)) is 1/5 ln|x-1|-1/5 ln|x+4|+C