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

The integral from 1 to 4 of 1/(x(1+sqrt(x))) is 2(2ln(2)-ln(3))

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

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

2 (2 ln (2) - ln (3))

Decimal

0.57536 \dots

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

\frac{1}{u ( u + 1 )} : \frac{1}{u} - \frac{1}{u + 1}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int_{1}^{2} \frac{1}{u} d u - \int_{1}^{2} \frac{1}{u + 1} d u)

\int_{1}^{2} \frac{1}{u} d u = ln (2)

\int_{1}^{2} \frac{1}{u + 1} d u = ln (3) - ln (2)

= 2 (ln (2) - (ln (3) - ln (2)))

Simplify

= 2 (2 ln (2) - ln (3))

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What is the integral from 1 to 4 of 1/(x(1+sqrt(x))) ?
The integral from 1 to 4 of 1/(x(1+sqrt(x))) is 2(2ln(2)-ln(3))