What is the integral of 1/(x^2sqrt(8-x^2)) ?

The integral of 1/(x^2sqrt(8-x^2)) is -(sqrt(8-x^2))/(8x)+C

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integral of 1/(x^2sqrt(8-x^2))

\int \frac{1}{x ^{2} 8 - x ^{2}} d x

- \frac{8 - x ^{2}}{8 x} + C

Solution steps

\int \frac{1}{x ^{2} 8 - x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{8 sin ( u ) cos ( u )} d u

= \int \frac{cot ( u )}{8 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}{8} \cdot \int \frac{cot ( u )}{sin ( u ) cos ( u )} d u

Simplify

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

= \frac{1}{8} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= \frac{1}{8} (- cot (u))

Substitute back

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

= \frac{1}{8} (- cot (arcsin (\frac{1}{2 2} x)))

Simplify

\frac{1}{8} (- cot (arcsin (\frac{1}{2 2} x))) : - \frac{8 - x ^{2}}{8 x}

= - \frac{8 - x ^{2}}{8 x}

Add a constant to the solution

= - \frac{8 - x ^{2}}{8 x} + C

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

\int \frac{x}{x + 3} d x

\frac{\partial}{\partial y} (x + e^{y})

t \to 3 lim (2 t^{- 1})

\frac{d}{d x} (\frac{5 + 6 x - 8 x}{x})

What is the integral of 1/(x^2sqrt(8-x^2)) ?
The integral of 1/(x^2sqrt(8-x^2)) is -(sqrt(8-x^2))/(8x)+C