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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \int \frac{1}{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}{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}{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

\frac{1 - 2 x ^{2}}{x ^{4} - 2 x ^{2} + 6}

h \to 0 lim (\frac{( f ( h ) ( 0 + h ) - f ( 0 ) )}{h})

\int x f (x) d x

f (x) = 17 e^{x}

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