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

The integral of (x+2)/(x^2+5x-6) is 2(1/4 ln|4x^2+20x-24|+1/28 (ln| 12/7+2/7 x|-ln|-2/7+2/7 x|))+C

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

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

2 (\frac{1}{4} ln 4 x^{2} + 20 x - 24 + \frac{1}{28} (ln \frac{12}{7} + \frac{2}{7} x - ln - \frac{2}{7} + \frac{2}{7} x)) + C

Solution steps

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

Complete the square

x^{2} + 5 x - 6 : (x + \frac{5}{2})^{2} - \frac{49}{4}

= \int \frac{x + 2}{( x + \frac{5}{2} ) ^{2} - \frac{49}{4}} d x

Apply u-substitution

: \int \frac{2 ( 2 u - 1 )}{4 u ^{2} - 49} d u

= \int \frac{2 ( 2 u - 1 )}{4 u ^{2} - 49} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{2 u - 1}{4 u ^{2} - 49} d u

Expand

\frac{2 u - 1}{4 u ^{2} - 49} : \frac{2 u}{4 u ^{2} - 49} - \frac{1}{4 u ^{2} - 49}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int \frac{2 u}{4 u ^{2} - 49} d u - \int \frac{1}{4 u ^{2} - 49} d u)

\int \frac{2 u}{4 u ^{2} - 49} d u = \frac{1}{4} ln 4 u^{2} - 49

\int \frac{1}{4 u ^{2} - 49} d u = - \frac{1}{28} (ln \frac{2}{7} u + 1 - ln \frac{2}{7} u - 1)

= 2 (\frac{1}{4} ln 4 u^{2} - 49 - (- \frac{1}{28} (ln \frac{2}{7} u + 1 - ln \frac{2}{7} u - 1)))

Substitute back

u = x + \frac{5}{2}

= 2 (\frac{1}{4} ln 4 (x + \frac{5}{2})^{2} - 49 - (- \frac{1}{28} (ln \frac{2}{7} (x + \frac{5}{2}) + 1 - ln \frac{2}{7} (x + \frac{5}{2}) - 1)))

Simplify

2 (\frac{1}{4} ln 4 (x + \frac{5}{2})^{2} - 49 - (- \frac{1}{28} (ln \frac{2}{7} (x + \frac{5}{2}) + 1 - ln \frac{2}{7} (x + \frac{5}{2}) - 1))) : 2 (\frac{1}{4} ln 4 x^{2} + 20 x - 24 + \frac{1}{28} (ln \frac{12}{7} + \frac{2}{7} x - ln - \frac{2}{7} + \frac{2}{7} x))

= 2 (\frac{1}{4} ln 4 x^{2} + 20 x - 24 + \frac{1}{28} (ln \frac{12}{7} + \frac{2}{7} x - ln - \frac{2}{7} + \frac{2}{7} x))

Add a constant to the solution

= 2 (\frac{1}{4} ln 4 x^{2} + 20 x - 24 + \frac{1}{28} (ln \frac{12}{7} + \frac{2}{7} x - ln - \frac{2}{7} + \frac{2}{7} x)) + C

\int_{0}^{1} x^{2} x d x

\frac{x}{x ^{2} - 7}

\int (x^{4} - x^{2} + \frac{1}{x ^{3}} - \frac{1}{x ^{2}}) d x

\int 3 + 2 x y d x

y = - 2 x - 4

What is the integral of (x+2)/(x^2+5x-6) ?
The integral of (x+2)/(x^2+5x-6) is 2(1/4 ln|4x^2+20x-24|+1/28 (ln| 12/7+2/7 x|-ln|-2/7+2/7 x|))+C