What is the integral from 2 to 4 of (sqrt(16-x^2))/x ?

The integral from 2 to 4 of (sqrt(16-x^2))/x is 2(ln((sqrt(3))/2+1)-sqrt(3)-ln(-(sqrt(3))/2+1))

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integral from 2 to 4 of (sqrt(16-x^2))/x

\int_{2}^{4} \frac{16 - x ^{2}}{x} d x

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

Decimal Notation

1.80372 \dots

Solution steps

\int_{2}^{4} \frac{16 - x ^{2}}{x} d x

Apply u-substitution

: \int_{23}^{0} - \frac{u ^{2}}{- u ^{2} + 16} d u

= \int_{23}^{0} - \frac{u ^{2}}{- u ^{2} + 16} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{23} - \frac{u ^{2}}{- u ^{2} + 16} d u

Take the constant out:

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

= - (- \int_{0}^{23} \frac{u ^{2}}{- u ^{2} + 16} d u)

Apply Integral Substitution

= - (- \int_{0}^{\frac{3}{2}} \frac{4 v ^{2}}{- v ^{2} + 1} d v)

Take the constant out:

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

= - (- 4 \cdot \int_{0}^{\frac{3}{2}} \frac{v ^{2}}{- v ^{2} + 1} d v)

\frac{v ^{2}}{- v ^{2} + 1} = \frac{1}{- v ^{2} + 1} - 1

= - (- 4 \cdot \int_{0}^{\frac{3}{2}} \frac{1}{- v ^{2} + 1} - 1 d v)

Apply the Sum Rule:

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

= - (- 4 (\int_{0}^{\frac{3}{2}} \frac{1}{- v ^{2} + 1} d v - \int_{0}^{\frac{3}{2}} 1 d v))

\int_{0}^{\frac{3}{2}} \frac{1}{- v ^{2} + 1} d v = \frac{1}{2} (ln (\frac{3}{2} + 1) - ln (- \frac{3}{2} + 1))

\int_{0}^{\frac{3}{2}} 1 d v = \frac{3}{2}

= - (- 4 (\frac{1}{2} (ln (\frac{3}{2} + 1) - ln (- \frac{3}{2} + 1)) - \frac{3}{2}))

Simplify

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

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

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What is the integral from 2 to 4 of (sqrt(16-x^2))/x ?
The integral from 2 to 4 of (sqrt(16-x^2))/x is 2(ln((sqrt(3))/2+1)-sqrt(3)-ln(-(sqrt(3))/2+1))