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

The integral from 0 to x of arctan(t^2) is

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integral from 0 to x of arctan(t^2)

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

x arctan (x^{2}) - 2 (\frac{1}{4 2} (2 arctan (2 x - 1) + ln 2 x^{2} - 22 x + 2) - \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1)))

Solution steps

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

Apply Integration By Parts

: [t arctan (t^{2}) - \int \frac{2 t ^{2}}{t ^{4} + 1} d t]_{0}^{x}

= [t arctan (t^{2}) - \int \frac{2 t ^{2}}{t ^{4} + 1} d t]_{0}^{x}

\int \frac{2 t ^{2}}{t ^{4} + 1} d t = 2 (- \frac{1}{4 2} (ln 2 t^{2} + 22 t + 2 - 2 arctan (2 t + 1)) + \frac{1}{4 2} (ln 2 t^{2} - 22 t + 2 + 2 arctan (2 t - 1)))

= [t arctan (t^{2}) - 2 (- \frac{1}{4 2} (ln 2 t^{2} + 22 t + 2 - 2 arctan (2 t + 1)) + \frac{1}{4 2} (ln 2 t^{2} - 22 t + 2 + 2 arctan (2 t - 1)))]_{0}^{x}

Compute the boundaries:

x arctan (x^{2}) - 2 (\frac{1}{4 2} (2 arctan (2 x - 1) + ln 2 x^{2} - 22 x + 2) - \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1)))

= x arctan (x^{2}) - 2 (\frac{1}{4 2} (2 arctan (2 x - 1) + ln 2 x^{2} - 22 x + 2) - \frac{1}{4 2} (ln 2 x^{2} + 22 x + 2 - 2 arctan (2 x + 1)))

\frac{d}{d x} (\frac{5}{4 x ^{3}})

\int \frac{1}{x - 1 + x + 1} d x

e^{x} \cdot cos (x)

\int e^{- x} ln (x) d x

y = 3 x^{2} - 5 x, (2, 2)

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