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

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

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

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

2 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) + C

Solution steps

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

Complete the square

x^{2} + x - 6 : (x + \frac{1}{2})^{2} - \frac{25}{4}

= \int \frac{x}{( x + \frac{1}{2} ) ^{2} - \frac{25}{4}} d x

Apply u-substitution

: \int \frac{2 ( 2 u - 1 )}{4 u ^{2} - 25} d u

= \int \frac{2 ( 2 u - 1 )}{4 u ^{2} - 25} 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} - 25} d u

Expand

\frac{2 u - 1}{4 u ^{2} - 25} : \frac{2 u}{4 u ^{2} - 25} - \frac{1}{4 u ^{2} - 25}

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} - 25} d u - \int \frac{1}{4 u ^{2} - 25} d u)

\int \frac{2 u}{4 u ^{2} - 25} d u = \frac{1}{4} ln 4 u^{2} - 25

\int \frac{1}{4 u ^{2} - 25} d u = - \frac{1}{20} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1)

= 2 (\frac{1}{4} ln 4 u^{2} - 25 - (- \frac{1}{20} (ln \frac{2}{5} u + 1 - ln \frac{2}{5} u - 1)))

Substitute back

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

= 2 (\frac{1}{4} ln 4 (x + \frac{1}{2})^{2} - 25 - (- \frac{1}{20} (ln \frac{2}{5} (x + \frac{1}{2}) + 1 - ln \frac{2}{5} (x + \frac{1}{2}) - 1)))

Simplify

2 (\frac{1}{4} ln 4 (x + \frac{1}{2})^{2} - 25 - (- \frac{1}{20} (ln \frac{2}{5} (x + \frac{1}{2}) + 1 - ln \frac{2}{5} (x + \frac{1}{2}) - 1))) : 2 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x))

= 2 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x))

Add a constant to the solution

= 2 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) + C

\frac{2}{3 ( s + 1 ) ^{2} - 27}

\int \frac{x + 5}{x ^{2} + 9} d x

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

y^{^{'}} + e^{x} y = 7 e^{x}

(7 x + 1) (x^{2} - 6)

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