What is the integral from 0 to 1 of x/(sqrt(1-x^2)) ?

The integral from 0 to 1 of x/(sqrt(1-x^2)) is 1

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integral from 0 to 1 of x/(sqrt(1-x^2))

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

1

Solution steps

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

Apply u-substitution

: \int_{1}^{0} - \frac{1}{2 u} d u

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

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

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

Take the constant out:

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

= - (- \frac{1}{2} \cdot \int_{0}^{1} \frac{1}{u} d u)

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{0}^{1}

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

Simplify

- (- \frac{1}{2} [2 u^{\frac{1}{2}}]_{0}^{1}) : \frac{1}{2} [2 u]_{0}^{1}

= \frac{1}{2} [2 u]_{0}^{1}

Compute the boundaries:

2

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

Simplify

= 1

\frac{x d y}{d x} = 5 x e^{x} - y + 6 x^{2}

\int (x^{5} - 4 x)^{2} (5 x^{4} - 4) d x

x \to 0 + lim (\frac{1}{1 + \frac{1}{x}})

\frac{d}{d x} (- 8)

x \to 0 lim (\frac{sin ^{2} ( 3 x )}{5 x ^{2}})

What is the integral from 0 to 1 of x/(sqrt(1-x^2)) ?
The integral from 0 to 1 of x/(sqrt(1-x^2)) is 1