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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

= \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

\frac{cot ( u )}{cos ( u ) sin ( u )} : csc^{2} (u)

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

Use the common integral:

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

= - cot (u)

Substitute back

u = arcsin (x)

= - cot (arcsin (x))

Simplify

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

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

Add a constant to the solution

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

5 x^{2} + 4 x + 2

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (a x)

2 x^{3} + 3 x^{2} - 12 x + 5

\frac{d y}{d x} - \frac{2 x y}{x ^{2} + 1} = 1

\frac{d ^{2} y}{d x ^{2}} - 49 y = e^{7 x}

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