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

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

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

\int \frac{x}{( 1 + x ^{4} )} d x

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

Solution steps

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

Apply u-substitution

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

= \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 = x^{2}

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

Add a constant to the solution

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

x \to - 3 lim ((6 - 3 x) (\frac{3}{5} x - 1))

\int 2 y^{2} + y e^{x y} d x

(5 y^{2} - x y) d x + (x^{2}) d y = 0

y^{^{''}} - 3 y^{^{'}} = 8

x \to 1 - lim ((x - 2)^{2})

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