Hey there! Let's walk through solving that remaining integral using trigonometric substitution step by step—you've already made great progress with integration by parts, so we just need to tie up the last piece.

First, let's recap where you're at:

You've used integration by parts to get to this point:
$$
\int \frac{\sqrt{36x^{2}-49}}{x{3}}dx = -\frac{\sqrt{36x^{2}-49}}{2x{2}}+\int\frac{18}{x\sqrt{36x^{2}-49}}dx
$$

The key here is tackling the integral ( \int\frac{18}{x\sqrt{36x^{2}-49}}dx ). Since we have a square root of the form ( \sqrt{a^2x2 - b^2} ) (here, ( 36x^2 = (6x)^2 ) and ( 49 = 7^2 )), a secant substitution is perfect—it leverages the trig identity ( \sec^2u - 1 = \tan^2u ) to eliminate the square root cleanly.

Let's break this down step by step:

1. Set up the trigonometric substitution
   Let ( 6x = 7\sec u ). This choice works because when we plug it into the square root:
   $$
   \sqrt{36x^2 - 49} = \sqrt{(7\sec u)^2 - 7^2} = 7\sqrt{\sec^2u - 1} = 7\tan u
   $$
   (Since ( x > \frac{7}{6} ), ( 6x > 7 ), so ( \sec u > 1 ), meaning ( u ) is in the interval ( (0, \frac{\pi}{2}) ) where ( \tan u ) is positive—no need for absolute values.)

2. Find ( dx ) in terms of ( du )
   Differentiate both sides of ( 6x = 7\sec u ):
   $$
   6dx = 7\sec u \tan u , du \implies dx = \frac{7\sec u \tan u}{6} du
   $$

3. Substitute into the integral and simplify
   First, solve for ( x ): ( x = \frac{7\sec u}{6} ). Now plug ( x ), ( \sqrt{36x^2 -49} ), and ( dx ) into the integral:
   $$
   18 \int \frac{1}{\left(\frac{7\sec u}{6}\right) \cdot 7\tan u} \cdot \frac{7\sec u \tan u}{6} du
   $$
   Let's cancel out terms step by step:

   • The ( 7\sec u \tan u ) in the numerator cancels with the ( 7\sec u \tan u ) in the denominator
   • The remaining coefficients simplify to ( 18 \times \frac{6}{7 \times 6} = \frac{18}{7} )

   So the integral simplifies to:
   $$
   \frac{18}{7} \int 1 , du = \frac{18}{7}u + C
   $$

4. Convert back to ( x )
   Since ( 6x = 7\sec u ), we can solve for ( u ):
   $$
   \sec u = \frac{6x}{7} \implies u = \text{arcsec}\left(\frac{6x}{7}\right)
   $$
   Alternatively, you can write this as ( u = \arccos\left(\frac{7}{6x}\right) ) (since ( \sec u = \frac{1}{\cos u} ))—both forms are valid.

Final Result #

Plug this back into your original integration by parts expression, and you get:
$$
\int \frac{\sqrt{36x^{2}-49}}{x{3}}dx = -\frac{\sqrt{36x^{2}-49}}{2x{2}}+\frac{18}{7}\text{arcsec}\left(\frac{6x}{7}\right)+C
$$

To remember this for future problems: anytime you see ( \sqrt{t^2 - a^2} ) in an integral, secant substitution is your go-to—it's designed to turn that square root into a tangent term, making the integral much easier to solve.

备注：内容来源于stack exchange，提问作者Atlas