What is the integral of-5/(x^2sqrt(5-x^2)) ?

The integral of-5/(x^2sqrt(5-x^2)) is (sqrt(5-x^2))/x+C

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integral of-5/(x^2sqrt(5-x^2))

\int - \frac{5}{x ^{2} 5 - x ^{2}} d x

\frac{5 - x ^{2}}{x} + C

Solution steps

\int - \frac{5}{x ^{2} 5 - x ^{2}} d x

Take the constant out:

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

= - 5 \cdot \int \frac{1}{x ^{2} 5 - x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{5 sin ( u ) cos ( u )} d u

= - 5 \cdot \int \frac{cot ( u )}{5 sin ( u ) cos ( u )} d u

Take the constant out:

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

= - 5 \cdot \frac{1}{5} \cdot \int \frac{cot ( u )}{sin ( u ) cos ( u )} d u

Simplify

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

= - 5 \cdot \frac{1}{5} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= - 5 \cdot \frac{1}{5} (- cot (u))

Substitute back

u = arcsin (\frac{1}{5} x)

= - 5 \cdot \frac{1}{5} (- cot (arcsin (\frac{1}{5} x)))

Simplify

- 5 \cdot \frac{1}{5} (- cot (arcsin (\frac{1}{5} x))) : \frac{5 - x ^{2}}{x}

= \frac{5 - x ^{2}}{x}

Add a constant to the solution

= \frac{5 - x ^{2}}{x} + C

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What is the integral of-5/(x^2sqrt(5-x^2)) ?
The integral of-5/(x^2sqrt(5-x^2)) is (sqrt(5-x^2))/x+C