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

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

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integral from 0 to 1 of arctan(1-x)

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

- \frac{1}{2} ln (2) + \frac{π}{4}

Decimal Notation

0.43882 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int - \frac{x}{x ^{2} - 2 x + 2} d x = - \frac{1}{2} ln x^{2} - 2 x + 2 - arctan (x - 1)

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

Simplify

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

Compute the boundaries:

- (\frac{1}{2} ln (2) - \frac{π}{4})

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

Simplify

= - \frac{1}{2} ln (2) + \frac{π}{4}

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\int_{1}^{5} \frac{18 ( ln ( x ) ) ^{5}}{x} d x

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