What is the integral from 0 to 1 of xsqrt(x/(1-x)) ?

The integral from 0 to 1 of xsqrt(x/(1-x)) is (3pi)/8

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integral from 0 to 1 of xsqrt(x/(1-x))

\int_{0}^{1} x \frac{x}{1 - x} d x

\frac{3 π}{8}

Decimal

1.17809 \dots

Solution steps

\int_{0}^{1} x \frac{x}{1 - x} d x

Apply u-substitution

: \int_{1}^{0} - \frac{- u + 1}{u} (- u + 1) d u

= \int_{1}^{0} - \frac{- u + 1}{u} (- u + 1) d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} - \frac{- u + 1}{u} (- u + 1) d u

Take the constant out:

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

= - (- \int_{0}^{1} \frac{- u + 1}{u} (- u + 1) d u)

Expand

\frac{- u + 1}{u} (- u + 1) : - \frac{- u + 1}{u} u + \frac{- u + 1}{u}

= - (- \int_{0}^{1} - \frac{- u + 1}{u} u + \frac{- u + 1}{u} d u)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- (- \int_{0}^{1} \frac{- u + 1}{u} u d u + \int_{0}^{1} \frac{- u + 1}{u} d u))

\int_{0}^{1} \frac{- u + 1}{u} u d u = \frac{π}{8}

\int_{0}^{1} \frac{- u + 1}{u} d u = \frac{π}{2}

= - (- (- \frac{π}{8} + \frac{π}{2}))

Simplify

- (- (- \frac{π}{8} + \frac{π}{2})) : \frac{3 π}{8}

= \frac{3 π}{8}

\int_{1}^{0} x - x^{2} d x

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

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\int_{0}^{22 π} sin (x) d x

\int_{0}^{x} (x + 8 y) d y

What is the integral from 0 to 1 of xsqrt(x/(1-x)) ?
The integral from 0 to 1 of xsqrt(x/(1-x)) is (3pi)/8