What is the integral of (sqrt(36x^2-49))/(x^3) ?

The integral of (sqrt(36x^2-49))/(x^3) is -(sqrt(36x^2-49))/(2x^2)+18/7 arcsec(6/7 x)+C

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integral of (sqrt(36x^2-49))/(x^3)

\int \frac{36 x ^{2} - 49}{x ^{3}} d x

- \frac{36 x ^{2} - 49}{2 x ^{2}} + \frac{18}{7} arcsec (\frac{6}{7} x) + C

Solution steps

\int \frac{36 x ^{2} - 49}{x ^{3}} d x

Apply Integration By Parts

: - \frac{36 x ^{2} - 49}{2 x ^{2}} - \int - \frac{18}{x 36 x ^{2} - 49} d x

= - \frac{36 x ^{2} - 49}{2 x ^{2}} - \int - \frac{18}{x 36 x ^{2} - 49} d x

\int - \frac{18}{x 36 x ^{2} - 49} d x = - \frac{18}{7} arcsec (\frac{6}{7} x)

= - \frac{36 x ^{2} - 49}{2 x ^{2}} - (- \frac{18}{7} arcsec (\frac{6}{7} x))

Simplify

= - \frac{36 x ^{2} - 49}{2 x ^{2}} + \frac{18}{7} arcsec (\frac{6}{7} x)

Add a constant to the solution

= - \frac{36 x ^{2} - 49}{2 x ^{2}} + \frac{18}{7} arcsec (\frac{6}{7} x) + C

x \to \frac{π}{2} - lim (sec (x))

\frac{d y}{d x} = e^{8 x - 3 y}, y (0) = 4

f (x) = \frac{x ^{2}}{( 4 + x )}

- 9 x^{\frac{1}{3}} + 5, at x = 27

\frac{d}{d x} (x^{2} - 2 x - \frac{2}{x ^{4}})

What is the integral of (sqrt(36x^2-49))/(x^3) ?
The integral of (sqrt(36x^2-49))/(x^3) is -(sqrt(36x^2-49))/(2x^2)+18/7 arcsec(6/7 x)+C