What is the integral of (sqrt(16x^2-49))/(x^3) ?

The integral of (sqrt(16x^2-49))/(x^3) is -(sqrt(16x^2-49))/(2x^2)+8/7 arcsec(4/7 x)+C

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integral of (sqrt(16x^2-49))/(x^3)

\int \frac{16 x ^{2} - 49}{x ^{3}} d x

- \frac{16 x ^{2} - 49}{2 x ^{2}} + \frac{8}{7} arcsec (\frac{4}{7} x) + C

Solution steps

\int \frac{16 x ^{2} - 49}{x ^{3}} d x

Apply Integration By Parts

: - \frac{16 x ^{2} - 49}{2 x ^{2}} - \int - \frac{8}{x 16 x ^{2} - 49} d x

= - \frac{16 x ^{2} - 49}{2 x ^{2}} - \int - \frac{8}{x 16 x ^{2} - 49} d x

\int - \frac{8}{x 16 x ^{2} - 49} d x = - \frac{8}{7} arcsec (\frac{4}{7} x)

= - \frac{16 x ^{2} - 49}{2 x ^{2}} - (- \frac{8}{7} arcsec (\frac{4}{7} x))

Simplify

= - \frac{16 x ^{2} - 49}{2 x ^{2}} + \frac{8}{7} arcsec (\frac{4}{7} x)

Add a constant to the solution

= - \frac{16 x ^{2} - 49}{2 x ^{2}} + \frac{8}{7} arcsec (\frac{4}{7} x) + C

\frac{( y - 3 )}{y ^{2} - 3 y + 9}

y^{^{''}} + 4 y = t - 2 sin (2 t)

\frac{d}{d y} (x^{2} y sin^{3} (x y^{2}))

x \to 0 lim (arcsin (e^{x}))

\frac{d}{d x} (\frac{( x - 1 )}{x + 1})

What is the integral of (sqrt(16x^2-49))/(x^3) ?
The integral of (sqrt(16x^2-49))/(x^3) is -(sqrt(16x^2-49))/(2x^2)+8/7 arcsec(4/7 x)+C