What is the integral of (-36)/(x^2sqrt(9-x^2)) ?

The integral of (-36)/(x^2sqrt(9-x^2)) is (4sqrt(-x^2+9))/x+C

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integral of (-36)/(x^2sqrt(9-x^2))

\int \frac{- 36}{x ^{2} 9 - x ^{2}} d x

\frac{4 - x ^{2} + 9}{x} + C

Solution steps

\int \frac{- 36}{x ^{2} 9 - x ^{2}} d x

Take the constant out:

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

= - 36 \cdot \int \frac{1}{x ^{2} 9 - x ^{2}} d x

Apply Trigonometric Substitution

= - 36 \cdot \int \frac{cot ( u )}{9 cos ( u ) sin ( u )} d u

Take the constant out:

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

= - 36 \cdot \frac{1}{9} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

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

= - 36 \cdot \frac{1}{9} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= - 36 \cdot \frac{1}{9} (- cot (u))

Substitute back

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

= - 36 \cdot \frac{1}{9} (- cot (arcsin (\frac{1}{3} x)))

Simplify

- 36 \cdot \frac{1}{9} (- cot (arcsin (\frac{1}{3} x))) : \frac{4 - x ^{2} + 9}{x}

= \frac{4 - x ^{2} + 9}{x}

Add a constant to the solution

= \frac{4 - x ^{2} + 9}{x} + C

What is the integral of (-36)/(x^2sqrt(9-x^2)) ?
The integral of (-36)/(x^2sqrt(9-x^2)) is (4sqrt(-x^2+9))/x+C