What is the integral from 0 to x of y/(1+y^4) ?

The integral from 0 to x of y/(1+y^4) is 1/2 arctan(x^2)

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integral from 0 to x of y/(1+y^4)

\int_{0}^{x} \frac{y}{1 + y ^{4}} d y

\frac{1}{2} arctan (x^{2})

Solution steps

\int_{0}^{x} \frac{y}{1 + y ^{4}} d y

Compute the indefinite integral:

\int \frac{y}{1 + y ^{4}} d y = \frac{1}{2} arctan (y^{2}) + C

Compute the boundaries:

\int_{0}^{x} \frac{y}{1 + y ^{4}} d y = \frac{1}{2} arctan (x^{2}) - 0

= \frac{1}{2} arctan (x^{2}) - 0

Simplify

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

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What is the integral from 0 to x of y/(1+y^4) ?
The integral from 0 to x of y/(1+y^4) is 1/2 arctan(x^2)