What is the integral from 1 to 3 of x/(x^4+1) ?

The integral from 1 to 3 of x/(x^4+1) is 1/2 arctan(9)-pi/8

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integral from 1 to 3 of x/(x^4+1)

\int_{1}^{3} \frac{x}{x ^{4} + 1} d x

\frac{1}{2} arctan (9) - \frac{π}{8}

Decimal Notation

0.33737 \dots

Solution steps

\int_{1}^{3} \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}^{3} \frac{x}{x ^{4} + 1} d x = \frac{1}{2} arctan (9) - \frac{π}{8}

= \frac{1}{2} arctan (9) - \frac{π}{8}

\int_{0}^{1} 2 x x^{2} + 3 d x

\int_{0}^{10} \frac{1}{20} d y

\int_{0}^{4} (\frac{1}{5}) e^{\frac{- x}{5}} d x

What is the integral from 1 to 3 of x/(x^4+1) ?
The integral from 1 to 3 of x/(x^4+1) is 1/2 arctan(9)-pi/8