What is the integral of 1/(2xsqrt(9-x^2)) ?

The integral of 1/(2xsqrt(9-x^2)) is -1/12 (ln|(sqrt(9-x^2))/3+1|-ln|(sqrt(9-x^2))/3-1|)+C

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integral of 1/(2xsqrt(9-x^2))

\int \frac{1}{2 x 9 - x ^{2}} d x

- \frac{1}{12} (ln \frac{9 - x ^{2}}{3} + 1 - ln \frac{9 - x ^{2}}{3} - 1) + C

Solution steps

\int \frac{1}{2 x 9 - x ^{2}} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{x 9 - x ^{2}} d x

Apply u-substitution

= \frac{1}{2} \cdot \int - \frac{1}{- u ^{2} + 9} d u

Take the constant out:

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

= \frac{1}{2} (- \int \frac{1}{- u ^{2} + 9} d u)

Apply Integral Substitution

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

Substitute back

= \frac{1}{2} - \frac{1}{3} \frac{ln \frac{9 - x ^{2}}{3} + 1}{2} - \frac{ln \frac{9 - x ^{2}}{3} - 1}{2}

Simplify

\frac{1}{2} - \frac{1}{3} \frac{ln \frac{9 - x ^{2}}{3} + 1}{2} - \frac{ln \frac{9 - x ^{2}}{3} - 1}{2} : - \frac{1}{12} (ln \frac{9 - x ^{2}}{3} + 1 - ln \frac{9 - x ^{2}}{3} - 1)

= - \frac{1}{12} (ln \frac{9 - x ^{2}}{3} + 1 - ln \frac{9 - x ^{2}}{3} - 1)

Add a constant to the solution

= - \frac{1}{12} (ln \frac{9 - x ^{2}}{3} + 1 - ln \frac{9 - x ^{2}}{3} - 1) + C

What is the integral of 1/(2xsqrt(9-x^2)) ?
The integral of 1/(2xsqrt(9-x^2)) is -1/12 (ln|(sqrt(9-x^2))/3+1|-ln|(sqrt(9-x^2))/3-1|)+C