What is the integral of 1/(xsqrt(36-x^2)) ?

The integral of 1/(xsqrt(36-x^2)) is -1/12 (ln|(sqrt(36-x^2))/6+1|-ln|(sqrt(36-x^2))/6-1|)+C

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integral of 1/(xsqrt(36-x^2))

\int \frac{1}{x 36 - x ^{2}} d x

- \frac{1}{12} (ln \frac{36 - x ^{2}}{6} + 1 - ln \frac{36 - x ^{2}}{6} - 1) + C

Solution steps

\int \frac{1}{x 36 - x ^{2}} d x

Apply u-substitution

: \int - \frac{1}{- u ^{2} + 36} d u

= \int - \frac{1}{- u ^{2} + 36} d u

Take the constant out:

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

= - \int \frac{1}{- u ^{2} + 36} d u

Apply Integral Substitution

= - \int \frac{1}{6 ( - 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}{6} \cdot \int \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}{6} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= - \frac{1}{6} \frac{ln \frac{36 - x ^{2}}{6} + 1}{2} - \frac{ln \frac{36 - x ^{2}}{6} - 1}{2}

Simplify

- \frac{1}{6} \frac{ln \frac{36 - x ^{2}}{6} + 1}{2} - \frac{ln \frac{36 - x ^{2}}{6} - 1}{2} : - \frac{1}{12} (ln \frac{36 - x ^{2}}{6} + 1 - ln \frac{36 - x ^{2}}{6} - 1)

= - \frac{1}{12} (ln \frac{36 - x ^{2}}{6} + 1 - ln \frac{36 - x ^{2}}{6} - 1)

Add a constant to the solution

= - \frac{1}{12} (ln \frac{36 - x ^{2}}{6} + 1 - ln \frac{36 - x ^{2}}{6} - 1) + C

x \to - π lim (cot (x))

\frac{1}{( s ^{2} ( s + 1 ) )}

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

\int \frac{6 x 2 x ^{2} - 7}{5} d x

\frac{d}{d x} (\frac{3}{2 x - 1})

What is the integral of 1/(xsqrt(36-x^2)) ?
The integral of 1/(xsqrt(36-x^2)) is -1/12 (ln|(sqrt(36-x^2))/6+1|-ln|(sqrt(36-x^2))/6-1|)+C