What is the integral of ((6x^2+7x-6))/((x^2-4)(x+2)) ?

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

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integral of ((6x^2+7x-6))/((x^2-4)(x+2))

\int \frac{( 6 x ^{2} + 7 x - 6 )}{( x ^{2} - 4 ) ( x + 2 )} d x

4 ln ∣ x + 2 ∣ + \frac{1}{x + 2} + 2 ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{6 x ^{2} + 7 x - 6}{( x ^{2} - 4 ) ( x + 2 )} d x

Take the partial fraction of

\frac{6 x ^{2} + 7 x - 6}{( x ^{2} - 4 ) ( x + 2 )} : \frac{4}{x + 2} - \frac{1}{( x + 2 ) ^{2}} + \frac{2}{x - 2}

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

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

\int \frac{4}{x + 2} d x = 4 ln ∣ x + 2 ∣

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

\int \frac{2}{x - 2} d x = 2 ln ∣ x - 2 ∣

= 4 ln ∣ x + 2 ∣ - (- \frac{1}{x + 2}) + 2 ln ∣ x - 2 ∣

Simplify

= 4 ln ∣ x + 2 ∣ + \frac{1}{x + 2} + 2 ln ∣ x - 2 ∣

Add a constant to the solution

= 4 ln ∣ x + 2 ∣ + \frac{1}{x + 2} + 2 ln ∣ x - 2 ∣ + C

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

\int_{0}^{4} 4 - x^{2} d x

\frac{d x}{d t} - 3 x = t

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

f (x) = x^{2} + 4, (- 5, 29)

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