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

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

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

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

2 π

Decimal Notation

6.28318 \dots

Solution steps

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

Apply u-substitution

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

= 4 \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

= 4 \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)

= 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

y = 4 (x + 1), y = 5 (x + 1), x = 2

\frac{d}{d x} (3 x^{2} y^{2})

\frac{d}{d x} (x (1 - x^{2})^{\frac{1}{2}})

n = 0 \sum \infty i (- 1)^{n}

(x + x) \frac{d y}{d x} = y + y

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