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

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

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integral of (2u^2-1)/(u^3-u)

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

ln ∣ u ∣ + \frac{1}{2} ln ∣ u + 1 ∣ + \frac{1}{2} ln ∣ u - 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{2 u ^{2} - 1}{u ^{3} - u} : \frac{1}{u} + \frac{1}{2 ( u + 1 )} + \frac{1}{2 ( u - 1 )}

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

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

= ln ∣ u ∣ + \frac{1}{2} ln ∣ u + 1 ∣ + \frac{1}{2} ln ∣ u - 1 ∣

Add a constant to the solution

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

\int_{100}^{120} 600 d x

\int \frac{8 x}{3 - 4 x ^{2}} d x

\int_{1}^{4} (3 x^{2} + 1) d x

4 x - \frac{5 y x ^{2} + 1 d y}{d x} = 0, y (0) = - 10

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