What is the integral of (sqrt(9x^2-49))/(x^3) ?

The integral of (sqrt(9x^2-49))/(x^3) is -(sqrt(9x^2-49))/(2x^2)+9/14 arcsec(3/7 x)+C

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integral of (sqrt(9x^2-49))/(x^3)

\int \frac{9 x ^{2} - 49}{x ^{3}} d x

- \frac{9 x ^{2} - 49}{2 x ^{2}} + \frac{9}{14} arcsec (\frac{3}{7} x) + C

Solution steps

\int \frac{9 x ^{2} - 49}{x ^{3}} d x

Apply Integration By Parts

: - \frac{9 x ^{2} - 49}{2 x ^{2}} - \int - \frac{9}{2 x 9 x ^{2} - 49} d x

= - \frac{9 x ^{2} - 49}{2 x ^{2}} - \int - \frac{9}{2 x 9 x ^{2} - 49} d x

\int - \frac{9}{2 x 9 x ^{2} - 49} d x = - \frac{9}{14} arcsec (\frac{3}{7} x)

= - \frac{9 x ^{2} - 49}{2 x ^{2}} - (- \frac{9}{14} arcsec (\frac{3}{7} x))

Simplify

= - \frac{9 x ^{2} - 49}{2 x ^{2}} + \frac{9}{14} arcsec (\frac{3}{7} x)

Add a constant to the solution

= - \frac{9 x ^{2} - 49}{2 x ^{2}} + \frac{9}{14} arcsec (\frac{3}{7} x) + C

7 (t - 7)

\int (x^{4} + x + 1 + x^{- 1} + x^{- 2}) d x

\int_{9}^{16} \frac{x}{x - 4} d x

f (x) = ∣ 4 + 7 x ∣

What is the integral of (sqrt(9x^2-49))/(x^3) ?
The integral of (sqrt(9x^2-49))/(x^3) is -(sqrt(9x^2-49))/(2x^2)+9/14 arcsec(3/7 x)+C