What is the integral of (sqrt(-9x^2+16))/(x^2) ?

The integral of (sqrt(-9x^2+16))/(x^2) is -(sqrt(-9x^2+16))/x-3arcsin(3/4 x)+C

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integral of (sqrt(-9x^2+16))/(x^2)

\int \frac{- 9 x ^{2} + 16}{x ^{2}} d x

- \frac{- 9 x ^{2} + 16}{x} - 3 arcsin (\frac{3}{4} x) + C

Solution steps

\int \frac{- 9 x ^{2} + 16}{x ^{2}} d x

Apply Integration By Parts

: - \frac{- 9 x ^{2} + 16}{x} - \int \frac{9}{- 9 x ^{2} + 16} d x

= - \frac{- 9 x ^{2} + 16}{x} - \int \frac{9}{- 9 x ^{2} + 16} d x

\int \frac{9}{- 9 x ^{2} + 16} d x = 3 arcsin (\frac{3}{4} x)

= - \frac{- 9 x ^{2} + 16}{x} - 3 arcsin (\frac{3}{4} x)

Add a constant to the solution

= - \frac{- 9 x ^{2} + 16}{x} - 3 arcsin (\frac{3}{4} x) + C

\int \frac{x + 1}{3 ( x ^{2} - x + 1 )} d x

\int \frac{1}{( 9 - 4 x ) 16 x ^{2} - 72 x + 80} d x

\int \frac{( t ^{2} + 1 ) e ^{t}}{( t + 1 ) ^{2}} d t

\int \frac{9 x ^{3} - 7 x + 2}{x ^{3} - x ^{2}} d x

\int x \cdot arccot (x) d x

What is the integral of (sqrt(-9x^2+16))/(x^2) ?
The integral of (sqrt(-9x^2+16))/(x^2) is -(sqrt(-9x^2+16))/x-3arcsin(3/4 x)+C