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

\int \frac{1}{25 x ^{2} + 36} d x

\frac{d}{d x} (m x^{m - 1})

\frac{( 2 s + 2 )}{s ^{2} + 2 s + 5}

(1 - y sin (x)) d x + cos (x) d y = 0

y = \frac{5}{x}

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