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 0 + lim (\frac{1}{2 + e ^{\frac{1}{x}}})

\int_{1}^{9} (x^{2} - 4 x + 5) d x

\frac{d}{d x} (x^{l o g_{e} (x)})

\frac{1}{3 + t}

x \to 3 lim (\frac{x ^{2} - 4}{x - 2})

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