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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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}{u ( u - 1 )} d u

Apply u-substitution

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

Factor

v^{2} - 1 : - (- v^{2} + 1)

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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