What is the integral of (x+3)/((x-2)(x+5)) ?

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

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integral of (x+3)/((x-2)(x+5))

\int \frac{x + 3}{( x - 2 ) ( x + 5 )} d x

ln ∣ x + 5 ∣ + 5 (\frac{1}{7} ln ∣ x - 2 ∣ - \frac{1}{7} ln ∣ x + 5 ∣) + C

Solution steps

\int \frac{x + 3}{( x - 2 ) ( x + 5 )} d x

Apply u-substitution

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

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

Expand

\frac{u + 5}{u ( u + 7 )} : \frac{u}{u ( u + 7 )} + \frac{5}{u ( u + 7 )}

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{u}{u ( u + 7 )} d u + \int \frac{5}{u ( u + 7 )} d u

\int \frac{u}{u ( u + 7 )} d u = ln ∣ u + 7 ∣

\int \frac{5}{u ( u + 7 )} d u = 5 (\frac{1}{7} ln ∣ u ∣ - \frac{1}{7} ln ∣ u + 7 ∣)

= ln ∣ u + 7 ∣ + 5 (\frac{1}{7} ln ∣ u ∣ - \frac{1}{7} ln ∣ u + 7 ∣)

Substitute back

u = x - 2

= ln ∣ x - 2 + 7 ∣ + 5 (\frac{1}{7} ln ∣ x - 2 ∣ - \frac{1}{7} ln ∣ x - 2 + 7 ∣)

Simplify

ln ∣ x - 2 + 7 ∣ + 5 (\frac{1}{7} ln ∣ x - 2 ∣ - \frac{1}{7} ln ∣ x - 2 + 7 ∣) : ln ∣ x + 5 ∣ + 5 (\frac{1}{7} ln ∣ x - 2 ∣ - \frac{1}{7} ln ∣ x + 5 ∣)

= ln ∣ x + 5 ∣ + 5 (\frac{1}{7} ln ∣ x - 2 ∣ - \frac{1}{7} ln ∣ x + 5 ∣)

Add a constant to the solution

= ln ∣ x + 5 ∣ + 5 (\frac{1}{7} ln ∣ x - 2 ∣ - \frac{1}{7} ln ∣ x + 5 ∣) + C

\frac{\partial}{\partial x} ((x - 1)^{\frac{2}{3}} (y^{2} - 4))

\frac{\partial}{\partial y} (y (x + y + 1))

\frac{d}{d x} ((x - 5)^{3})

\int \frac{1}{( x ^{2} - 2 x + 2 ) ^{2}} d x

\int \frac{4}{1 - x ^{2}} d x

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