What is the integral of (x-17)/(x^2+x-12) ?

The integral of (x-17)/(x^2+x-12) is 2(1/4 ln|4x^2+4x-48|+5/4 (ln| 8/7+2/7 x|-ln|-6/7+2/7 x|))+C

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integral of (x-17)/(x^2+x-12)

\int \frac{x - 17}{x ^{2} + x - 12} d x

2 (\frac{1}{4} ln 4 x^{2} + 4 x - 48 + \frac{5}{4} (ln \frac{8}{7} + \frac{2}{7} x - ln - \frac{6}{7} + \frac{2}{7} x)) + C

Solution steps

\int \frac{x - 17}{x ^{2} + x - 12} d x

Complete the square

x^{2} + x - 12 : (x + \frac{1}{2})^{2} - \frac{49}{4}

= \int \frac{x - 17}{( x + \frac{1}{2} ) ^{2} - \frac{49}{4}} d x

Apply u-substitution

: \int \frac{2 ( 2 u - 35 )}{4 u ^{2} - 49} d u

= \int \frac{2 ( 2 u - 35 )}{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 - 35}{4 u ^{2} - 49} d u

Expand

\frac{2 u - 35}{4 u ^{2} - 49} : \frac{2 u}{4 u ^{2} - 49} - \frac{35}{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{35}{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{35}{4 u ^{2} - 49} d u = - \frac{5}{4} (ln \frac{2}{7} u + 1 - ln \frac{2}{7} u - 1)

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

Substitute back

u = x + \frac{1}{2}

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

Simplify

2 (\frac{1}{4} ln 4 (x + \frac{1}{2})^{2} - 49 - (- \frac{5}{4} (ln \frac{2}{7} (x + \frac{1}{2}) + 1 - ln \frac{2}{7} (x + \frac{1}{2}) - 1))) : 2 (\frac{1}{4} ln 4 x^{2} + 4 x - 48 + \frac{5}{4} (ln \frac{8}{7} + \frac{2}{7} x - ln - \frac{6}{7} + \frac{2}{7} x))

= 2 (\frac{1}{4} ln 4 x^{2} + 4 x - 48 + \frac{5}{4} (ln \frac{8}{7} + \frac{2}{7} x - ln - \frac{6}{7} + \frac{2}{7} x))

Add a constant to the solution

= 2 (\frac{1}{4} ln 4 x^{2} + 4 x - 48 + \frac{5}{4} (ln \frac{8}{7} + \frac{2}{7} x - ln - \frac{6}{7} + \frac{2}{7} x)) + C

\frac{d}{d y} (\frac{1 + sin ( y )}{1 - sin ( y )})

\int 20 e^{- 0.6 t} d t

\int_{0}^{4} (2 x + 3) d x

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

f (x) = (x^{3} - 9 x)^{12}, at x = 3

What is the integral of (x-17)/(x^2+x-12) ?
The integral of (x-17)/(x^2+x-12) is 2(1/4 ln|4x^2+4x-48|+5/4 (ln| 8/7+2/7 x|-ln|-6/7+2/7 x|))+C