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

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

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

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

\frac{π}{2}

Decimal

1.57079 \dots

Solution steps

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

Apply u-substitution

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

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

Compute the boundaries:

\frac{π}{4}

= 2 \cdot \frac{π}{4}

Simplify

= \frac{π}{2}

\int y y + 1 d y

y = 5 x - x^{2}

4 tan^{2} (3 x)

\int - ln (x) d x

\frac{\partial}{\partial y} (\frac{1}{x ^{2} + y ^{2}})

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