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

y = \frac{1}{x ^{6}}, (1, 1)

\int \frac{4 x}{( 2 x ^{2} - 7 ) ^{10}} d x

3 t^{3}

\int \frac{1}{a ^{2} + u ^{2}} d u

(9, 1), (- 1, 1)

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