What is the integral of 1/(x^4+y^2) ?

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

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integral of 1/(x^4+y^2)

\int \frac{1}{x ^{4} + y ^{2}} d y

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

Solution steps

\int \frac{1}{x ^{4} + y ^{2}} d y

Apply Integral Substitution

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

Take the constant out:

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

= \frac{1}{x ^{2}} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

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

Substitute back

u = \frac{y}{x ^{2}}

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

Add a constant to the solution

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

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