What is the integral of 1/(x^2sqrt(9-4x^2)) ?

The integral of 1/(x^2sqrt(9-4x^2)) is -(sqrt(9-4x^2))/(9x)+C

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integral of 1/(x^2sqrt(9-4x^2))

\int \frac{1}{x ^{2} 9 - 4 x ^{2}} d x

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

Solution steps

\int \frac{1}{x ^{2} 9 - 4 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{2}{9 sin ^{2} ( u )} d u

= \int \frac{2}{9 sin ^{2} ( u )} d u

Take the constant out:

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

= \frac{2}{9} \cdot \int \frac{1}{sin ^{2} ( u )} d u

Use the common integral:

\int \frac{1}{sin ^{2} ( u )} d u = - cot (u)

= \frac{2}{9} (- cot (u))

Substitute back

u = arcsin (\frac{2}{3} x)

= \frac{2}{9} (- cot (arcsin (\frac{2}{3} x)))

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d y} (\frac{x}{x + y})

8 x - x^{2} (1.7)

x \to 0 lim (\frac{x}{- 2 x ^{5} + 15 x})

\int (1 - x) e^{x} d x

x \to \frac{1}{2} + lim (\frac{x}{2 x - 1})

What is the integral of 1/(x^2sqrt(9-4x^2)) ?
The integral of 1/(x^2sqrt(9-4x^2)) is -(sqrt(9-4x^2))/(9x)+C