What is the integral of (7x+3)/((x+4)(x-1)) ?

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

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integral of (7x+3)/((x+4)(x-1))

\int \frac{7 x + 3}{( x + 4 ) ( x - 1 )} d x

5 ln ∣ x + 4 ∣ + 2 ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{7 x + 3}{( x + 4 ) ( x - 1 )} d x

Apply u-substitution

: \int \frac{7 u - 25}{u ( u - 5 )} d u

= \int \frac{7 u - 25}{u ( u - 5 )} d u

Take the partial fraction of

\frac{7 u - 25}{u ( u - 5 )} : \frac{5}{u} + \frac{2}{u - 5}

= \int \frac{5}{u} + \frac{2}{u - 5} d u

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}{u} d u + \int \frac{2}{u - 5} d u

\int \frac{5}{u} d u = 5 ln ∣ u ∣

\int \frac{2}{u - 5} d u = 2 ln ∣ u - 5 ∣

= 5 ln ∣ u ∣ + 2 ln ∣ u - 5 ∣

Substitute back

u = x + 4

= 5 ln ∣ x + 4 ∣ + 2 ln ∣ x + 4 - 5 ∣

Simplify

= 5 ln ∣ x + 4 ∣ + 2 ln ∣ x - 1 ∣

Add a constant to the solution

= 5 ln ∣ x + 4 ∣ + 2 ln ∣ x - 1 ∣ + C

x \to 1 lim (\frac{\frac{1}{x} - 1}{1 - x})

x \to 0 lim (\frac{18}{x ^{2}})

\frac{d}{d x} (\frac{9}{( x - 8 ) ^{2}})

\frac{\partial}{\partial x} (3 x^{2} - 24 y)

\frac{d}{d x} (- 2 ln (x))

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