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

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

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

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

π

Decimal Notation

3.14159 \dots

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{1 + u ^{2}} 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

f (x) = arcsin (x)

\frac{2}{4 s ^{2} - 8 s}

\int_{2}^{4} \frac{( 2 x - 3 ) x}{6} d x

x \to - 2 lim (- x^{2} + 6 x - 1)

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

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