What is the integral from 1 to t^2 of arctan(u) ?

The integral from 1 to t^2 of arctan(u) is t^2arctan(t^2)-1/2 ln|t^4+1|-pi/4+1/2 ln(2)

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integral from 1 to t^2 of arctan(u)

\int_{1}^{t^{2}} arctan (u) d u

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

Solution steps

\int_{1}^{t^{2}} arctan (u) d u

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

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

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

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What is the integral from 1 to t^2 of arctan(u) ?
The integral from 1 to t^2 of arctan(u) is t^2arctan(t^2)-1/2 ln|t^4+1|-pi/4+1/2 ln(2)