What is the integral of (sqrt(4x^2-81))/(x^3) ?

The integral of (sqrt(4x^2-81))/(x^3) is -(sqrt(4x^2-81))/(2x^2)+2/9 arcsec(2/9 x)+C

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integral of (sqrt(4x^2-81))/(x^3)

\int \frac{4 x ^{2} - 81}{x ^{3}} d x

- \frac{4 x ^{2} - 81}{2 x ^{2}} + \frac{2}{9} arcsec (\frac{2}{9} x) + C

Solution steps

\int \frac{4 x ^{2} - 81}{x ^{3}} d x

Apply Integration By Parts

: - \frac{4 x ^{2} - 81}{2 x ^{2}} - \int - \frac{2}{x 4 x ^{2} - 81} d x

= - \frac{4 x ^{2} - 81}{2 x ^{2}} - \int - \frac{2}{x 4 x ^{2} - 81} d x

\int - \frac{2}{x 4 x ^{2} - 81} d x = - \frac{2}{9} arcsec (\frac{2}{9} x)

= - \frac{4 x ^{2} - 81}{2 x ^{2}} - (- \frac{2}{9} arcsec (\frac{2}{9} x))

Simplify

= - \frac{4 x ^{2} - 81}{2 x ^{2}} + \frac{2}{9} arcsec (\frac{2}{9} x)

Add a constant to the solution

= - \frac{4 x ^{2} - 81}{2 x ^{2}} + \frac{2}{9} arcsec (\frac{2}{9} x) + C

\frac{d}{d x} (\frac{3}{8 ( x + 1 ) ^{\frac{5}{2}}})

(e^{x})^{^{'}}

\int \frac{9}{x ( x ^{4} + 7 )} d x

\int 2 x^{2} cos (x^{3}) d x

f (x) = \frac{1}{x ^{3} + 2 x + 1}

What is the integral of (sqrt(4x^2-81))/(x^3) ?
The integral of (sqrt(4x^2-81))/(x^3) is -(sqrt(4x^2-81))/(2x^2)+2/9 arcsec(2/9 x)+C