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

The integral from 0 to 1 of xarctan(sqrt(x)) is 1/3

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integral from 0 to 1 of xarctan(sqrt(x))

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

\frac{1}{3}

Decimal

0.33333 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{1} 2 u^{3} arctan (u) d u

= \int_{0}^{1} 2 u^{3} arctan (u) d u

Take the constant out:

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

= 2 \cdot \int_{0}^{1} u^{3} arctan (u) d u

Apply Integration By Parts

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

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

\int \frac{u ^{4}}{4 ( u ^{2} + 1 )} d u = \frac{1}{4} (\frac{u ^{3}}{3} - u + arctan (u))

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

Compute the boundaries:

\frac{1}{6}

= 2 \cdot \frac{1}{6}

Simplify

= \frac{1}{3}

\int_{0}^{1} x^{2} x^{2} + 3 d x

\int_{0}^{1} \frac{1}{x ^{3} + x ^{2}} d x

\int_{1}^{4} \frac{6}{x} d x

\int_{- 3}^{4} e^{x + 7} d x

\int_{- 2}^{2} x^{3} - 3 x + 0.5 d x

What is the integral from 0 to 1 of xarctan(sqrt(x)) ?
The integral from 0 to 1 of xarctan(sqrt(x)) is 1/3