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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{4 cos ( u ) sin ( u )} d u

= \int \frac{cot ( u )}{4 cos ( u ) sin ( u )} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

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

= \frac{1}{4} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= \frac{1}{4} (- cot (u))

Substitute back

u = arcsin (\frac{1}{2} x)

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

Simplify

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

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

Add a constant to the solution

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

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

\int 2 x - sin (x) d x

\int sinh (a x) d x

\int (\frac{8}{1 + x ^{2}}) d x

\int (2 x + 3) (x^{2} + 3 x)^{4} d x

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