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

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

Popular >

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \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

y^{^{'}} = \frac{3}{3 y}

\int x^{3} ln (x^{2} + 1) d x

\int \frac{8}{t ^{9}} sin (\frac{1}{t ^{8}} - 9) d t

y^{^{'}} + tan (x) y = cos^{2} (x)

\frac{\partial}{\partial x} (tan (2 x - y))

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