What is the integral of (x^{-2})/(sqrt((2-x)(2+x))) ?

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

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integral of (x^{-2})/(sqrt((2-x)(2+x)))

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

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

Solution steps

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

Multiply by the conjugate of

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

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

Simplify

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

= \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

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