What is the integral of 8/(x^2sqrt(4-x^2)) ?

The integral of 8/(x^2sqrt(4-x^2)) is -(2sqrt(-x^2+4))/x+C

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integral of 8/(x^2sqrt(4-x^2))

\int \frac{8}{x ^{2} 4 - x ^{2}} d x

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

Solution steps

\int \frac{8}{x ^{2} 4 - x ^{2}} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{x ^{2} 4 - x ^{2}} d x

Apply Trigonometric Substitution

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

= 8 \cdot \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

= 8 \cdot \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)

= 8 \cdot \frac{1}{4} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= 8 \cdot \frac{1}{4} (- cot (u))

Substitute back

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

= 8 \cdot \frac{1}{4} (- cot (arcsin (\frac{1}{2} x)))

Simplify

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

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

Add a constant to the solution

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

\int csc (v) d v

y = sin^{3} (x)

\frac{\partial}{\partial x} (\frac{b}{2 b - k ^{2} - 1 - 2 k \cdot x})

\int 2 sin^{2} (x co) s^{2} x d x

7 sec (x) tan (x)

What is the integral of 8/(x^2sqrt(4-x^2)) ?
The integral of 8/(x^2sqrt(4-x^2)) is -(2sqrt(-x^2+4))/x+C