What is the integral of (-1-u)/(u^3+u^2-8u-12) ?

The integral of (-1-u)/(u^3+u^2-8u-12) is 4/25 ln|u+2|+1/(5(u+2))-4/25 ln|u-3|+C

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integral of (-1-u)/(u^3+u^2-8u-12)

\int \frac{- 1 - u}{u ^{3} + u ^{2} - 8 u - 12} d u

\frac{4}{25} ln ∣ u + 2 ∣ + \frac{1}{5 ( u + 2 )} - \frac{4}{25} ln ∣ u - 3 ∣ + C

Solution steps

\int \frac{- 1 - u}{u ^{3} + u ^{2} - 8 u - 12} d u

Take the partial fraction of

\frac{- 1 - u}{u ^{3} + u ^{2} - 8 u - 12} : \frac{4}{25 ( u + 2 )} - \frac{1}{5 ( u + 2 ) ^{2}} - \frac{4}{25 ( u - 3 )}

= \int \frac{4}{25 ( u + 2 )} - \frac{1}{5 ( u + 2 ) ^{2}} - \frac{4}{25 ( u - 3 )} d u

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}{25 ( u + 2 )} d u - \int \frac{1}{5 ( u + 2 ) ^{2}} d u - \int \frac{4}{25 ( u - 3 )} d u

\int \frac{4}{25 ( u + 2 )} d u = \frac{4}{25} ln ∣ u + 2 ∣

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

\int \frac{4}{25 ( u - 3 )} d u = \frac{4}{25} ln ∣ u - 3 ∣

= \frac{4}{25} ln ∣ u + 2 ∣ - (- \frac{1}{5 ( u + 2 )}) - \frac{4}{25} ln ∣ u - 3 ∣

Simplify

= \frac{4}{25} ln ∣ u + 2 ∣ + \frac{1}{5 ( u + 2 )} - \frac{4}{25} ln ∣ u - 3 ∣

Add a constant to the solution

= \frac{4}{25} ln ∣ u + 2 ∣ + \frac{1}{5 ( u + 2 )} - \frac{4}{25} ln ∣ u - 3 ∣ + C

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

\frac{d}{d x} (- \frac{13}{x ^{4}})

\frac{d}{d x} f (x) = \frac{12}{( 3 x - 4 ) ^{5}}

\int \frac{- 2}{z ^{3}} d z

\frac{d}{d x} ((3 x - y)^{2})

What is the integral of (-1-u)/(u^3+u^2-8u-12) ?
The integral of (-1-u)/(u^3+u^2-8u-12) is 4/25 ln|u+2|+1/(5(u+2))-4/25 ln|u-3|+C