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

The integral of (sqrt(9-4x^2))/(x^2) is -(sqrt(9-4x^2))/x-2arcsin(2/3 x)+C

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

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

- \frac{9 - 4 x ^{2}}{x} - 2 arcsin (\frac{2}{3} x) + C

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{4}{9 - 4 x ^{2}} d x = 2 arcsin (\frac{2}{3} x)

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

Add a constant to the solution

= - \frac{9 - 4 x ^{2}}{x} - 2 arcsin (\frac{2}{3} x) + C

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What is the integral of (sqrt(9-4x^2))/(x^2) ?
The integral of (sqrt(9-4x^2))/(x^2) is -(sqrt(9-4x^2))/x-2arcsin(2/3 x)+C