What is the integral of (2v)/(v^2-b^4) ?

The integral of (2v)/(v^2-b^4) is ln|-b^4+v^2|+C

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integral of (2v)/(v^2-b^4)

\int \frac{2 v}{v ^{2} - b ^{4}} d v

ln - b^{4} + v^{2} + C

Solution steps

\int \frac{2 v}{v ^{2} - b ^{4}} d v

Take the constant out:

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

= 2 \cdot \int \frac{v}{- b ^{4} + v ^{2}} d v

Apply u-substitution

: \int \frac{1}{2 u} d u

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = - b^{4} + v^{2}

= 2 \cdot \frac{1}{2} ln - b^{4} + v^{2}

Simplify

2 \cdot \frac{1}{2} ln - b^{4} + v^{2} : ln - b^{4} + v^{2}

= ln - b^{4} + v^{2}

Add a constant to the solution

= ln - b^{4} + v^{2} + C

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What is the integral of (2v)/(v^2-b^4) ?
The integral of (2v)/(v^2-b^4) is ln|-b^4+v^2|+C