What is the integral of (v^3)/(v^4+1) ?

The integral of (v^3)/(v^4+1) is 1/4 ln|v^4+1|+C

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integral of (v^3)/(v^4+1)

\int \frac{v ^{3}}{v ^{4} + 1} d v

\frac{1}{4} ln v^{4} + 1 + C

Solution steps

\int \frac{v ^{3}}{v ^{4} + 1} d v

Apply u-substitution

: \int \frac{1}{4 u} d u

= \int \frac{1}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = v^{4} + 1

= \frac{1}{4} ln v^{4} + 1

Add a constant to the solution

= \frac{1}{4} ln v^{4} + 1 + C

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What is the integral of (v^3)/(v^4+1) ?
The integral of (v^3)/(v^4+1) is 1/4 ln|v^4+1|+C