What is the integral from 0 to 2x of arctan(t) ?

The integral from 0 to 2x of arctan(t) is 2xarctan(2x)-1/2 ln|4x^2+1|

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integral from 0 to 2x of arctan(t)

\int_{0}^{2 x} arctan (t) d t

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

Solution steps

\int_{0}^{2 x} arctan (t) d t

Apply Integration By Parts

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

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

Compute the boundaries:

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

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

\int_{0}^{9} \frac{1}{3 + t} d t

\int_{\frac{1}{2}}^{\frac{e}{2}} t ln (2 t) d t

\int_{0}^{- 5} (\frac{2}{5} x^{2} + 2 x - 2) d x

\int_{0}^{20} 2 + 2 t d t

\int_{0}^{1} (4 x - x^{2}) d x

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