What is the integral from 0 to 1 of 6/(sqrt(x)(x+1)) ?

The integral from 0 to 1 of 6/(sqrt(x)(x+1)) is 3pi

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integral from 0 to 1 of 6/(sqrt(x)(x+1))

\int_{0}^{1} \frac{6}{x ( x + 1 )} d x

3 π

Solution steps

\int_{0}^{1} \frac{6}{x ( x + 1 )} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{1} \frac{1}{x ( x + 1 )} d x

Apply u-substitution

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

= 6 \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

= 6 \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)

= 6 \cdot 2 [arctan (u)]_{0}^{1}

Simplify

= 12 [arctan (u)]_{0}^{1}

Compute the boundaries:

\frac{π}{4}

= 12 \cdot \frac{π}{4}

Simplify

= 3 π

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What is the integral from 0 to 1 of 6/(sqrt(x)(x+1)) ?
The integral from 0 to 1 of 6/(sqrt(x)(x+1)) is 3pi