What is the integral from 0 to 1 of 2xarctan(x^2) ?

The integral from 0 to 1 of 2xarctan(x^2) is 2(pi/8-1/4 ln(2))

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integral from 0 to 1 of 2xarctan(x^2)

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

2 (\frac{π}{8} - \frac{1}{4} ln (2))

Decimal Notation

0.43882 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{1} x arctan (x^{2}) d x

Apply Integration By Parts

: [\frac{1}{2} x^{2} arctan (x^{2}) - \int \frac{x ^{3}}{x ^{4} + 1} d x]_{0}^{1}

= 2 [\frac{1}{2} x^{2} arctan (x^{2}) - \int \frac{x ^{3}}{x ^{4} + 1} d x]_{0}^{1}

\int \frac{x ^{3}}{x ^{4} + 1} d x = \frac{1}{4} ln x^{4} + 1

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

Compute the boundaries:

\frac{π}{8} - \frac{1}{4} ln (2)

= 2 (\frac{π}{8} - \frac{1}{4} ln (2))

\frac{d y}{d x} = 4 x^{5} y - y

\int_{- 2}^{0} (\frac{y}{2} + y + 3) d y

e^{- 2 t} sin (5 t)

6 e^{x^{- 5}}

f (x) = \frac{1}{3 + 2 x}, (1, \frac{1}{5})

What is the integral from 0 to 1 of 2xarctan(x^2) ?
The integral from 0 to 1 of 2xarctan(x^2) is 2(pi/8-1/4 ln(2))