What is the integral of (2x+1)/(x^3-7x+6) ?

The integral of (2x+1)/(x^3-7x+6) is -3/4 ln|x-1|+ln|x-2|-1/4 ln|x+3|+C

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integral of (2x+1)/(x^3-7x+6)

\int \frac{2 x + 1}{x ^{3} - 7 x + 6} d x

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

Solution steps

\int \frac{2 x + 1}{x ^{3} - 7 x + 6} d x

Take the partial fraction of

\frac{2 x + 1}{x ^{3} - 7 x + 6} : - \frac{3}{4 ( x - 1 )} + \frac{1}{x - 2} - \frac{1}{4 ( x + 3 )}

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

\int \frac{3}{4 ( x - 1 )} d x = \frac{3}{4} ln ∣ x - 1 ∣

\int \frac{1}{x - 2} d x = ln ∣ x - 2 ∣

\int \frac{1}{4 ( x + 3 )} d x = \frac{1}{4} ln ∣ x + 3 ∣

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

Add a constant to the solution

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

y^{^{'}} = \frac{( y ^{2} + 2 )}{x y - 4 y}

n = 0 \sum \infty (\frac{3}{4})^{2}

\int (x - 3)^{2} d x

\int 7 tan^{3} (x) d x

1300 + 14 x - 0.1 x^{2} + 0.0005 x^{3}

What is the integral of (2x+1)/(x^3-7x+6) ?
The integral of (2x+1)/(x^3-7x+6) is -3/4 ln|x-1|+ln|x-2|-1/4 ln|x+3|+C