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

The integral from 1 to infinity of x/(x^4+1) is pi/8

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integral from 1 to infinity of x/(x^4+1)

\int_{1}^{\infty} \frac{x}{x ^{4} + 1} d x

\frac{π}{8}

Decimal Notation

0.39269 \dots

Solution steps

\int_{1}^{\infty} \frac{x}{x ^{4} + 1} d x

Compute the indefinite integral:

\int \frac{x}{x ^{4} + 1} d x = \frac{1}{2} arctan (x^{2}) + C

Compute the boundaries:

\int_{1}^{\infty} \frac{x}{x ^{4} + 1} d x = \frac{π}{4} - \frac{π}{8}

= \frac{π}{4} - \frac{π}{8}

Simplify

= \frac{π}{8}

y = \frac{1 + 2 x}{3 + x}, (2, 1)

\frac{\partial}{\partial x} (5 x^{2} + 4 y^{2})

\int (x^{4} + 1) e^{x^{5} + 5 x} d x

n = 1 \sum \infty \frac{1}{( 2 n ) !}

x \to 0 lim ((1 - 6 x)^{\frac{1}{x}})

What is the integral from 1 to infinity of x/(x^4+1) ?
The integral from 1 to infinity of x/(x^4+1) is pi/8