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

The integral of 1/(2xsqrt(1-4x^2)) is -1/4 (ln|sqrt(1-4x^2)+1|-ln|sqrt(1-4x^2)-1|)+C

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

\int \frac{1}{2 x 1 - 4 x ^{2}} d x

- \frac{1}{4} (ln 1 - 4 x^{2} + 1 - ln 1 - 4 x^{2} - 1) + C

Solution steps

\int \frac{1}{2 x 1 - 4 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 1 - 4 x ^{2}} d x

Apply u-substitution

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

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

Factor

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

= \frac{1}{2} \cdot \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

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

Substitute back

u = 1 - 4 x^{2}

= \frac{1}{2} (- (\frac{ln 1 - 4 x ^{2} + 1}{2} - \frac{ln 1 - 4 x ^{2} - 1}{2}))

Simplify

\frac{1}{2} (- (\frac{ln 1 - 4 x ^{2} + 1}{2} - \frac{ln 1 - 4 x ^{2} - 1}{2})) : - \frac{1}{4} (ln 1 - 4 x^{2} + 1 - ln 1 - 4 x^{2} - 1)

= - \frac{1}{4} (ln 1 - 4 x^{2} + 1 - ln 1 - 4 x^{2} - 1)

Add a constant to the solution

= - \frac{1}{4} (ln 1 - 4 x^{2} + 1 - ln 1 - 4 x^{2} - 1) + C

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What is the integral of 1/(2xsqrt(1-4x^2)) ?
The integral of 1/(2xsqrt(1-4x^2)) is -1/4 (ln|sqrt(1-4x^2)+1|-ln|sqrt(1-4x^2)-1|)+C