What is the integral of (sqrt(16x^2-25))/(x^3) ?

The integral of (sqrt(16x^2-25))/(x^3) is -(sqrt(16x^2-25))/(2x^2)+8/5 arcsec(4/5 x)+C

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integral of (sqrt(16x^2-25))/(x^3)

\int \frac{16 x ^{2} - 25}{x ^{3}} d x

- \frac{16 x ^{2} - 25}{2 x ^{2}} + \frac{8}{5} arcsec (\frac{4}{5} x) + C

Solution steps

\int \frac{16 x ^{2} - 25}{x ^{3}} d x

Apply Integration By Parts

: - \frac{16 x ^{2} - 25}{2 x ^{2}} - \int - \frac{8}{x 16 x ^{2} - 25} d x

= - \frac{16 x ^{2} - 25}{2 x ^{2}} - \int - \frac{8}{x 16 x ^{2} - 25} d x

\int - \frac{8}{x 16 x ^{2} - 25} d x = - \frac{8}{5} arcsec (\frac{4}{5} x)

= - \frac{16 x ^{2} - 25}{2 x ^{2}} - (- \frac{8}{5} arcsec (\frac{4}{5} x))

Simplify

= - \frac{16 x ^{2} - 25}{2 x ^{2}} + \frac{8}{5} arcsec (\frac{4}{5} x)

Add a constant to the solution

= - \frac{16 x ^{2} - 25}{2 x ^{2}} + \frac{8}{5} arcsec (\frac{4}{5} x) + C

f (x) = 7 x + 2

\int \frac{5}{3 x + 1} d x

y^{^{''}} - 8 y^{^{'}} + 10 y = 0, y (0) = 0, y^{^{'}} (0) = 5

n = 1 \sum \infty \frac{1}{2 n !}

\int - 1 + ln (x) d x

What is the integral of (sqrt(16x^2-25))/(x^3) ?
The integral of (sqrt(16x^2-25))/(x^3) is -(sqrt(16x^2-25))/(2x^2)+8/5 arcsec(4/5 x)+C