What is the integral of (3x+5)/(x^3-x^2-x+1) ?

The integral of (3x+5)/(x^3-x^2-x+1) is -1/2 ln|x-1|-4/(x-1)+1/2 ln|x+1|+C

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integral of (3x+5)/(x^3-x^2-x+1)

\int \frac{3 x + 5}{x ^{3} - x ^{2} - x + 1} d x

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

Solution steps

\int \frac{3 x + 5}{x ^{3} - x ^{2} - x + 1} d x

Factor

\frac{3 x + 5}{x ^{3} - x ^{2} - x + 1}

= \int \frac{3 x + 5}{( x - 1 ) ^{2} ( x + 1 )} d x

Take the partial fraction of

\frac{3 x + 5}{( x - 1 ) ^{2} ( x + 1 )} : - \frac{1}{2 ( x - 1 )} + \frac{4}{( x - 1 ) ^{2}} + \frac{1}{2 ( x + 1 )}

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

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

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

\int \frac{1}{2 ( x + 1 )} d x = \frac{1}{2} ln ∣ x + 1 ∣

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

Add a constant to the solution

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

\frac{d}{d x} (arctan (sin (x^{2} + \frac{π}{x})))

y^{^{''}} + y^{^{'}} + y = t

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

y^{^{''}} + 5 y = 0, y^{^{'}} (1) = 5

- \frac{1}{4 x ^{\frac{3}{2}}}

What is the integral of (3x+5)/(x^3-x^2-x+1) ?
The integral of (3x+5)/(x^3-x^2-x+1) is -1/2 ln|x-1|-4/(x-1)+1/2 ln|x+1|+C