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

The integral of (-3x+2)/(x^3+x^2-2) is -1/5 ln|x-1|+1/10 (ln|x^2+2x+2|-26arctan(x+1))+C

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

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

- \frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{10} (ln x^{2} + 2 x + 2 - 26 arctan (x + 1)) + C

Solution steps

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

Take the partial fraction of

\frac{- 3 x + 2}{x ^{3} + x ^{2} - 2} : - \frac{1}{5 ( x - 1 )} + \frac{x - 12}{5 ( x ^{2} + 2 x + 2 )}

= \int - \frac{1}{5 ( x - 1 )} + \frac{x - 12}{5 ( x ^{2} + 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{1}{5 ( x - 1 )} d x + \int \frac{x - 12}{5 ( x ^{2} + 2 x + 2 )} d x

\int \frac{1}{5 ( x - 1 )} d x = \frac{1}{5} ln ∣ x - 1 ∣

\int \frac{x - 12}{5 ( x ^{2} + 2 x + 2 )} d x = \frac{1}{10} (ln x^{2} + 2 x + 2 - 26 arctan (x + 1))

= - \frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{10} (ln x^{2} + 2 x + 2 - 26 arctan (x + 1))

Add a constant to the solution

= - \frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{10} (ln x^{2} + 2 x + 2 - 26 arctan (x + 1)) + C

\int (\frac{3}{u ^{2}} + \frac{4}{u ^{3}} - 2) d u

y^{^{''}} - 4 y^{^{'}} = sin (3 x)

7 x^{6} - 4 x^{5}

\int cos (2 x) + sin (x) d x

\int 10000 - x^{2} d x

What is the integral of (-3x+2)/(x^3+x^2-2) ?
The integral of (-3x+2)/(x^3+x^2-2) is -1/5 ln|x-1|+1/10 (ln|x^2+2x+2|-26arctan(x+1))+C