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

The integral from 0 to x of arctan(t) is xarctan(x)-1/2 ln|x^2+1|

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integral from 0 to x of arctan(t)

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

Compute the boundaries:

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

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

\int_{0}^{π} 0 d x

\int_{0}^{\frac{π}{4}} 4 sec^{2} (x) d x

\int_{0}^{\frac{π}{2}} sin^{3} (θ) d θ

\int_{0}^{x} e^{x} d x

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

What is the integral from 0 to x of arctan(t) ?
The integral from 0 to x of arctan(t) is xarctan(x)-1/2 ln|x^2+1|