What is the integral of ((x^2-7))/(x^3-12x+16) ?

The integral of ((x^2-7))/(x^3-12x+16) is 3/4 ln|x-2|+1/(2(x-2))+1/4 ln|x+4|+C

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integral of ((x^2-7))/(x^3-12x+16)

\int \frac{( x ^{2} - 7 )}{x ^{3} - 12 x + 16} d x

\frac{3}{4} ln ∣ x - 2 ∣ + \frac{1}{2 ( x - 2 )} + \frac{1}{4} ln ∣ x + 4 ∣ + C

Solution steps

\int \frac{x ^{2} - 7}{x ^{3} - 12 x + 16} d x

Take the partial fraction of

\frac{x ^{2} - 7}{x ^{3} - 12 x + 16} : \frac{3}{4 ( x - 2 )} - \frac{1}{2 ( x - 2 ) ^{2}} + \frac{1}{4 ( x + 4 )}

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

\int \frac{3}{4 ( x - 2 )} d x = \frac{3}{4} ln ∣ x - 2 ∣

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

\int \frac{1}{4 ( x + 4 )} d x = \frac{1}{4} ln ∣ x + 4 ∣

= \frac{3}{4} ln ∣ x - 2 ∣ - (- \frac{1}{2 ( x - 2 )}) + \frac{1}{4} ln ∣ x + 4 ∣

Simplify

= \frac{3}{4} ln ∣ x - 2 ∣ + \frac{1}{2 ( x - 2 )} + \frac{1}{4} ln ∣ x + 4 ∣

Add a constant to the solution

= \frac{3}{4} ln ∣ x - 2 ∣ + \frac{1}{2 ( x - 2 )} + \frac{1}{4} ln ∣ x + 4 ∣ + C

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What is the integral of ((x^2-7))/(x^3-12x+16) ?
The integral of ((x^2-7))/(x^3-12x+16) is 3/4 ln|x-2|+1/(2(x-2))+1/4 ln|x+4|+C