What is the integral of (sqrt(49-16x^2))/(x^2) ?

The integral of (sqrt(49-16x^2))/(x^2) is -(sqrt(49-16x^2))/x-4arcsin(4/7 x)+C

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integral of (sqrt(49-16x^2))/(x^2)

\int \frac{49 - 16 x ^{2}}{x ^{2}} d x

- \frac{49 - 16 x ^{2}}{x} - 4 arcsin (\frac{4}{7} x) + C

Solution steps

\int \frac{49 - 16 x ^{2}}{x ^{2}} d x

Apply Integration By Parts

: - \frac{49 - 16 x ^{2}}{x} - \int \frac{16}{49 - 16 x ^{2}} d x

= - \frac{49 - 16 x ^{2}}{x} - \int \frac{16}{49 - 16 x ^{2}} d x

\int \frac{16}{49 - 16 x ^{2}} d x = 4 arcsin (\frac{4}{7} x)

= - \frac{49 - 16 x ^{2}}{x} - 4 arcsin (\frac{4}{7} x)

Add a constant to the solution

= - \frac{49 - 16 x ^{2}}{x} - 4 arcsin (\frac{4}{7} x) + C

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What is the integral of (sqrt(49-16x^2))/(x^2) ?
The integral of (sqrt(49-16x^2))/(x^2) is -(sqrt(49-16x^2))/x-4arcsin(4/7 x)+C