What is the integral from 1 to 2 of 1/(xsqrt(4-x^2)) ?

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

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integral from 1 to 2 of 1/(xsqrt(4-x^2))

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

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

Decimal

0.65847 \dots

Solution steps

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

Apply u-substitution

: \int_{3}^{0} - \frac{1}{- u ^{2} + 4} d u

= \int_{3}^{0} - \frac{1}{- u ^{2} + 4} d u

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

= - \int_{0}^{3} - \frac{1}{- u ^{2} + 4} d u

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= - (- \frac{1}{2} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{\frac{3}{2}})

Simplify

- (- \frac{1}{2} [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{\frac{3}{2}}) : \frac{1}{2} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{0}^{\frac{3}{2}}

= \frac{1}{2} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{0}^{\frac{3}{2}}

Compute the boundaries:

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

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

Simplify

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

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