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

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

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integral from 0 to 1 of arctan(y)

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

\frac{π}{4} - \frac{1}{2} ln (2)

Solution steps

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

\frac{π}{4} - \frac{1}{2} ln (2)

= \frac{π}{4} - \frac{1}{2} ln (2)

\int_{0}^{1} 8 x (x^{2} + 1)^{3} d x

\int_{2}^{7} x x + 2 d x

\int_{0}^{3} x^{2} + 1 d x

\int_{0}^{1} \frac{1}{x + 2} d x

\int_{1}^{3} (x + 3)^{3} d x

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