What is the integral from 0 to pi/4 of arctan(x) ?

The integral from 0 to pi/4 of arctan(x) is 1/2 (pi/2 arctan(pi/4)-ln((pi^2)/(16)+1))

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integral from 0 to pi/4 of arctan(x)

\int_{0}^{\frac{π}{4}} arctan (x) d x

\frac{1}{2} (\frac{π}{2} arctan (\frac{π}{4}) - ln (\frac{π ^{2}}{16} + 1))

Decimal

0.28265 \dots

Solution steps

\int_{0}^{\frac{π}{4}} arctan (x) d x

Apply Integration By Parts

: [x arctan (x) - \int \frac{x}{x ^{2} + 1} d x]_{0}^{\frac{π}{4}}

= [x arctan (x) - \int \frac{x}{x ^{2} + 1} d x]_{0}^{\frac{π}{4}}

\int \frac{x}{x ^{2} + 1} d x = \frac{1}{2} ln x^{2} + 1

= [x arctan (x) - \frac{1}{2} ln x^{2} + 1]_{0}^{\frac{π}{4}}

Simplify

[x arctan (x) - \frac{1}{2} ln x^{2} + 1]_{0}^{\frac{π}{4}} : [\frac{1}{2} (2 x arctan (x) - ln x^{2} + 1)]_{0}^{\frac{π}{4}}

= [\frac{1}{2} (2 x arctan (x) - ln x^{2} + 1)]_{0}^{\frac{π}{4}}

Compute the boundaries:

\frac{1}{2} (\frac{π}{2} arctan (\frac{π}{4}) - ln (\frac{π ^{2}}{16} + 1))

= \frac{1}{2} (\frac{π}{2} arctan (\frac{π}{4}) - ln (\frac{π ^{2}}{16} + 1))

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What is the integral from 0 to pi/4 of arctan(x) ?
The integral from 0 to pi/4 of arctan(x) is 1/2 (pi/2 arctan(pi/4)-ln((pi^2)/(16)+1))