What is the integral from 1 to 2 of x/(1+x^4) ?

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

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integral from 1 to 2 of x/(1+x^4)

\int_{1}^{2} \frac{x}{1 + x ^{4}} d x

\frac{1}{2} arctan (4) - \frac{π}{8}

Solution steps

\int_{1}^{2} \frac{x}{1 + x ^{4}} d x

Compute the indefinite integral:

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

Compute the boundaries:

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

= \frac{1}{2} arctan (4) - \frac{π}{8}

\int_{- a}^{a} x^{2} a^{2} - x^{2} d x

\int_{n}^{n + 1} \frac{1}{1 + x ^{2}} d x

\int_{\frac{π}{6}}^{\frac{π}{3}} csc^{2} (θ) d θ

\int_{0}^{1} (x - x^{2}) x d x

\int_{0}^{\frac{1}{2}} 2 x e^{- x^{2}} d x

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