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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int 2 csc^{2} (u) d u

= \int 2 csc^{2} (u) d u

Take the constant out:

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

= 2 \cdot \int csc^{2} (u) d u

Use the common integral:

\int csc^{2} (u) d u = - cot (u)

= 2 (- cot (u))

Substitute back

u = arcsin (2 x)

= 2 (- cot (arcsin (2 x)))

Simplify

2 (- cot (arcsin (2 x))) : - \frac{1 - 4 x ^{2}}{x}

= - \frac{1 - 4 x ^{2}}{x}

Add a constant to the solution

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

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