What is the integral from 0 to 1 of 4/(sqrt(x)(1+x)) ?

The integral from 0 to 1 of 4/(sqrt(x)(1+x)) is 2pi

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integral from 0 to 1 of 4/(sqrt(x)(1+x))

\int_{0}^{1} \frac{4}{x ( 1 + x )} d x

2 π

Decimal Notation

6.28318 \dots

Solution steps

\int_{0}^{1} \frac{4}{x ( 1 + x )} d x

Take the constant out:

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

= 4 \cdot \int_{0}^{1} \frac{1}{x ( 1 + x )} d x

Apply u-substitution

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

= 4 \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

= 4 \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)

= 4 \cdot 2 [arctan (u)]_{0}^{1}

Simplify

= 8 [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= 8 \cdot \frac{π}{4}

Simplify

= 2 π

θ \to 0 lim (\frac{2 sin ( 3 θ )}{θ})

\frac{d}{d x} (\frac{7}{x ln ( x )})

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

\frac{\partial}{\partial x} (a)

f (x) = (\frac{( x - 3 )}{( x + 2 )})^{2} x = - 1

What is the integral from 0 to 1 of 4/(sqrt(x)(1+x)) ?
The integral from 0 to 1 of 4/(sqrt(x)(1+x)) is 2pi