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

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

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integral from 1 to 4 of 1/(x(2+sqrt(x)))

\int_{1}^{4} \frac{1}{x ( 2 + x )} d x

- ln (2) + ln (3)

Decimal Notation

0.40546 \dots

Solution steps

\int_{1}^{4} \frac{1}{x ( 2 + x )} d x

Apply u-substitution

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

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

Take the partial fraction of

\frac{1}{u ( u + 2 )} : \frac{1}{2 u} - \frac{1}{2 ( u + 2 )}

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

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

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

= 2 (\frac{1}{2} ln (2) - (ln (2) - \frac{1}{2} ln (3)))

Simplify

2 (\frac{1}{2} ln (2) - (ln (2) - \frac{1}{2} ln (3))) : - ln (2) + ln (3)

= - ln (2) + ln (3)

\int_{3}^{4} \frac{e ^{l n (x)}}{x} d x

\int_{- 1}^{4} \frac{2 x ^{3}}{x ^{2} - 4} d x

\int_{2}^{4} 2 x^{3} + 2 x d x

\int_{1}^{3} x^{2} + x^{4} d x

\int_{- 1}^{0} - (x + 1)^{2} d x

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