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

The integral from 0 to 1 of 2/(sqrt(x)(x+1)) is pi

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

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

π

Decimal Notation

3.14159 \dots

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

= 2 \cdot 2 [arctan (u)]_{0}^{1}

Simplify

= 4 [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= 4 \cdot \frac{π}{4}

Simplify

= π

y = 4.2 x - x^{2}

\int \frac{700 x}{( x + 1 ) ^{\frac{1}{3}}} d x

\frac{\partial}{\partial x} (x e^{- x^{2} - y^{2}})

x \to 0 lim (\frac{e ^{x} - 1}{9 x})

\frac{d y}{d x} = 6 x^{2} + 3

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