What is the integral of (y^3)/(1+9y^4) ?

The integral of (y^3)/(1+9y^4) is 1/36 ln|1+9y^4|+C

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integral of (y^3)/(1+9y^4)

\int \frac{y ^{3}}{1 + 9 y ^{4}} d y

\frac{1}{36} ln 1 + 9 y^{4} + C

Solution steps

\int \frac{y ^{3}}{1 + 9 y ^{4}} d y

Apply u-substitution

: \int \frac{1}{36 u} d u

= \int \frac{1}{36 u} d u

Take the constant out:

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

= \frac{1}{36} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{36} ln ∣ u ∣

Substitute back

u = 1 + 9 y^{4}

= \frac{1}{36} ln 1 + 9 y^{4}

Add a constant to the solution

= \frac{1}{36} ln 1 + 9 y^{4} + C

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What is the integral of (y^3)/(1+9y^4) ?
The integral of (y^3)/(1+9y^4) is 1/36 ln|1+9y^4|+C