What is the integral from 0 to 1 of 1/(xsqrt(4-x^2)) ?

The integral from 0 to 1 of 1/(xsqrt(4-x^2)) is diverges

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integral from 0 to 1 of 1/(xsqrt(4-x^2))

\int_{0}^{1} \frac{1}{x 4 - x ^{2}} d x

diverges

Solution steps

\int_{0}^{1} \frac{1}{x 4 - x ^{2}} d x

Apply u-substitution

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

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

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

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

Apply Integral Substitution

= - (- \int_{\frac{3}{2}}^{1} \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_{\frac{3}{2}}^{1} \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}]_{\frac{3}{2}}^{1})

Simplify

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

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

Compute the boundaries:

diverges

= diverges

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

\int_{0}^{4} 2 x (x^{2} + 4)^{2} d x

\int_{0}^{3} \frac{1}{16 + x ^{2}} d x

\int_{- 1}^{0} (6 x) d x

\int_{2}^{5} \frac{( 5 x + 2 )}{3} d x

What is the integral from 0 to 1 of 1/(xsqrt(4-x^2)) ?
The integral from 0 to 1 of 1/(xsqrt(4-x^2)) is diverges