What is the integral of (v(v^2-1))/(v^2+1) ?

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

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integral of (v(v^2-1))/(v^2+1)

\int \frac{v ( v ^{2} - 1 )}{v ^{2} + 1} d v

- ln v^{2} + 1 + \frac{1}{2} v^{2} + C

Solution steps

\int \frac{v ( v ^{2} - 1 )}{v ^{2} + 1} d v

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

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

= \frac{1}{2} \cdot \int - \frac{2}{u + 1} + 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

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

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

\int 1 d u = u

= \frac{1}{2} (- 2 ln ∣ u + 1 ∣ + u)

Substitute back

u = v^{2}

= \frac{1}{2} (- 2 ln v^{2} + 1 + v^{2})

Simplify

\frac{1}{2} (- 2 ln v^{2} + 1 + v^{2}) : - ln v^{2} + 1 + \frac{1}{2} v^{2}

= - ln v^{2} + 1 + \frac{1}{2} v^{2}

Add a constant to the solution

= - ln v^{2} + 1 + \frac{1}{2} v^{2} + C

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