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

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

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integral from 0 to 1 of arctan(x)

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

\frac{π}{4} - \frac{1}{2} ln (2)

Decimal Notation

0.43882 \dots

Solution steps

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

Apply Integration By Parts

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

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

\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}^{1}

Compute the boundaries:

\frac{π}{4} - \frac{1}{2} ln (2)

= \frac{π}{4} - \frac{1}{2} ln (2)

\frac{\partial}{\partial x} ((x - 2)^{2} + (y - 2)^{2})

\int e^{x} + e d x

y = - 2 t^{2} + 3 t - 6

\frac{\partial}{\partial s} (r + s)

\int \frac{tan ( x )}{sec ( x )} d x

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