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

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

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

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

\frac{1}{2} ln 1 - x^{2} + 1 - \frac{1}{2} ln 1 - x^{2} - 1 + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Use the common integral:

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

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

Substitute back

u = 1 - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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