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

The integral of y/(1+y^4) is 1/2 arctan(y^2)+C

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integral of y/(1+y^4)

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

\frac{1}{2} arctan (y^{2}) + C

Solution steps

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

Apply u-substitution

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

Use the common integral:

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

= \frac{1}{2} arctan (u)

Substitute back

u = y^{2}

= \frac{1}{2} arctan (y^{2})

Add a constant to the solution

= \frac{1}{2} arctan (y^{2}) + C

What is the integral of y/(1+y^4) ?
The integral of y/(1+y^4) is 1/2 arctan(y^2)+C