What is the integral of (x+3)/(x^2+5x-6) ?

The integral of (x+3)/(x^2+5x-6) is ln|x+6|+4(1/7 ln|x-1|-1/7 ln|x+6|)+C

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integral of (x+3)/(x^2+5x-6)

\int \frac{x + 3}{x ^{2} + 5 x - 6} d x

ln ∣ x + 6 ∣ + 4 (\frac{1}{7} ln ∣ x - 1 ∣ - \frac{1}{7} ln ∣ x + 6 ∣) + C

Solution steps

\int \frac{x + 3}{x ^{2} + 5 x - 6} d x

Factor

x^{2} + 5 x - 6 : (x - 1) (x + 6)

= \int \frac{x + 3}{( x - 1 ) ( x + 6 )} d x

Apply u-substitution

: \int \frac{u + 4}{u ( u + 7 )} d u

= \int \frac{u + 4}{u ( u + 7 )} d u

Expand

\frac{u + 4}{u ( u + 7 )} : \frac{u}{u ( u + 7 )} + \frac{4}{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{4}{u ( u + 7 )} d u

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

\int \frac{4}{u ( u + 7 )} d u = 4 (\frac{1}{7} ln ∣ u ∣ - \frac{1}{7} ln ∣ u + 7 ∣)

= ln ∣ u + 7 ∣ + 4 (\frac{1}{7} ln ∣ u ∣ - \frac{1}{7} ln ∣ u + 7 ∣)

Substitute back

u = x - 1

= ln ∣ x - 1 + 7 ∣ + 4 (\frac{1}{7} ln ∣ x - 1 ∣ - \frac{1}{7} ln ∣ x - 1 + 7 ∣)

Simplify

ln ∣ x - 1 + 7 ∣ + 4 (\frac{1}{7} ln ∣ x - 1 ∣ - \frac{1}{7} ln ∣ x - 1 + 7 ∣) : ln ∣ x + 6 ∣ + 4 (\frac{1}{7} ln ∣ x - 1 ∣ - \frac{1}{7} ln ∣ x + 6 ∣)

= ln ∣ x + 6 ∣ + 4 (\frac{1}{7} ln ∣ x - 1 ∣ - \frac{1}{7} ln ∣ x + 6 ∣)

Add a constant to the solution

= ln ∣ x + 6 ∣ + 4 (\frac{1}{7} ln ∣ x - 1 ∣ - \frac{1}{7} ln ∣ x + 6 ∣) + C

x \to 1 lim (- 4)

2 x^{0}

\frac{d y}{d x} = y cos^{2} (y)

g (x) = x

x \to 3 lim ((3 x + 4)^{3})

What is the integral of (x+3)/(x^2+5x-6) ?
The integral of (x+3)/(x^2+5x-6) is ln|x+6|+4(1/7 ln|x-1|-1/7 ln|x+6|)+C