What is the integral of (-7sqrt(25x^2-9))/(x^3) ?

The integral of (-7sqrt(25x^2-9))/(x^3) is -7(-(sqrt(25x^2-9))/(2x^2)+25/6 arcsec(5/3 x))+C

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integral of (-7sqrt(25x^2-9))/(x^3)

\int \frac{- 7 25 x ^{2} - 9}{x ^{3}} d x

- 7 (- \frac{25 x ^{2} - 9}{2 x ^{2}} + \frac{25}{6} arcsec (\frac{5}{3} x)) + C

Solution steps

\int \frac{- 7 25 x ^{2} - 9}{x ^{3}} d x

Take the constant out:

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

= - 7 \cdot \int \frac{25 x ^{2} - 9}{x ^{3}} d x

Apply Integration By Parts

: - \frac{25 x ^{2} - 9}{2 x ^{2}} - \int - \frac{25}{2 x 25 x ^{2} - 9} d x

= - 7 (- \frac{25 x ^{2} - 9}{2 x ^{2}} - \int - \frac{25}{2 x 25 x ^{2} - 9} d x)

\int - \frac{25}{2 x 25 x ^{2} - 9} d x = - \frac{25}{6} arcsec (\frac{5}{3} x)

= - 7 (- \frac{25 x ^{2} - 9}{2 x ^{2}} - (- \frac{25}{6} arcsec (\frac{5}{3} x)))

Simplify

= - 7 (- \frac{25 x ^{2} - 9}{2 x ^{2}} + \frac{25}{6} arcsec (\frac{5}{3} x))

Add a constant to the solution

= - 7 (- \frac{25 x ^{2} - 9}{2 x ^{2}} + \frac{25}{6} arcsec (\frac{5}{3} x)) + C

x \to 0 lim ((1 + 2 x)^{\frac{3}{x}})

y^{^{''}} - 7 y^{^{'}} + 14 y = - 784 t^{2}

y^{^{''}} + 9 y = 3 tan (3 x)

\frac{x ^{2} - 6 x + 1}{x ^{2} + 9}

What is the integral of (-7sqrt(25x^2-9))/(x^3) ?
The integral of (-7sqrt(25x^2-9))/(x^3) is -7(-(sqrt(25x^2-9))/(2x^2)+25/6 arcsec(5/3 x))+C