What is the integral from 0 to 1/2 of arctan(x) ?

The integral from 0 to 1/2 of arctan(x) is (arctan(1/2)-ln(5/4))/2

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integral from 0 to 1/2 of arctan(x)

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

\frac{arctan ( \frac{1}{2} ) - ln ( \frac{5}{4} )}{2}

Decimal

0.12025 \dots

Solution steps

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

Apply Integration By Parts

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

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

\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{1}{2}}

Simplify

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

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

Compute the boundaries:

\frac{arctan ( \frac{1}{2} ) - ln ( \frac{5}{4} )}{2}

= \frac{arctan ( \frac{1}{2} ) - ln ( \frac{5}{4} )}{2}

\int_{0}^{1} x^{\frac{3}{8}} d x

\int_{0}^{\infty} e^{- α x^{2}} d x

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\int_{0}^{3} \frac{1}{3} x d x

\int_{0}^{\frac{π}{2}} 2 x d x

What is the integral from 0 to 1/2 of arctan(x) ?
The integral from 0 to 1/2 of arctan(x) is (arctan(1/2)-ln(5/4))/2