Integrating: Find \int\frac{5dx}{\sqrt{25x^2 -9}} x > \frac{3}{5}

[temaire]

Homework Statement

Find the integral of $$\int\frac{5dx}{\sqrt{25x^2 -9}}, x > \frac{3}{5}$$

The Attempt at a Solution

First, I made x = 3/5 secx, and dx = 3/5 secxtanxdx

$$\int\frac{3secxtanxdx}{5\sqrt{(9/25)(sec^2x -1)}}$$

$$\int\frac{secxtanxdx}{tanx}$$

$$\int secxdx$$

$$ln|secx + tanx| + C$$

$$ln|\frac{5x}{3} + \frac{5\sqrt{x^2 - \frac{9}{25}}}{3}| + C$$

The final step is my answer. However, when I try to integrate using the wolfram integration calculator, I get $$ln|2(\sqrt{25x^2 - 9} + 5x) + C$$

Where did I go wrong?

 

[rock.freak667]

Well I can't find an error in what you did but what I can say is that their answer can be reduced to


ln2+ln|5x+√(25x²-9)|+C=ln|5x+√(25x²-9)|+A


and your answer can be written as

ln(1/3)+ln|5x+√(25x²-9)| = ln|5x+√(25x²-9)|+B

So I would say that they are the same in essence.

 

[Mark44]

First, I made x = 3/5 secx, and dx = 3/5 secxtanxdx

It's not a good idea to have a substitution variable with the same name as the variable it is a substitution for.