What is the integral of 1/(x^2sqrt(3-x^2)) ?

The integral of 1/(x^2sqrt(3-x^2)) is -(sqrt(3-x^2))/(3x)+C

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integral of 1/(x^2sqrt(3-x^2))

\int \frac{1}{x ^{2} 3 - x ^{2}} d x

- \frac{3 - x ^{2}}{3 x} + C

Solution steps

\int \frac{1}{x ^{2} 3 - x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{3 sin ( u ) cos ( u )} d u

= \int \frac{cot ( u )}{3 sin ( u ) cos ( u )} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{cot ( u )}{sin ( u ) cos ( u )} d u

Simplify

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

= \frac{1}{3} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= \frac{1}{3} (- cot (u))

Substitute back

u = arcsin (\frac{1}{3} x)

= \frac{1}{3} (- cot (arcsin (\frac{1}{3} x)))

Simplify

\frac{1}{3} (- cot (arcsin (\frac{1}{3} x))) : - \frac{3 - x ^{2}}{3 x}

= - \frac{3 - x ^{2}}{3 x}

Add a constant to the solution

= - \frac{3 - x ^{2}}{3 x} + C

x \to \infty lim (- 2 x)

\frac{2 x ^{4} + 1}{y} \frac{d y}{d x} + \frac{1}{6} y^{3} x^{3} = 0

\frac{1}{2 x}

\int π x^{2} d x

\int x^{- 3} sin (x) d x

What is the integral of 1/(x^2sqrt(3-x^2)) ?
The integral of 1/(x^2sqrt(3-x^2)) is -(sqrt(3-x^2))/(3x)+C