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

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

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

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

2 (arctan (2) - \frac{π}{4})

Decimal

0.64350 \dots

Solution steps

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

Apply u-substitution

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

= \int_{1}^{2} \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

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

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= 2 [arctan (u)]_{1}^{2}

Compute the boundaries:

arctan (2) - \frac{π}{4}

= 2 (arctan (2) - \frac{π}{4})

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