Trouble seeing the difference between autograder answer and my own

[ago01]

Homework Statement

Integrate $\frac{1}{\sqrt{x^2-49}}dx$

Relevant Equations

Trig sub, secant.

I was doing a practice quiz and got the following integral:

$\int \frac{1}{\sqrt{x^2-49}}dx$

Following normal trig sub. Using $b^2 = c^2 - a^2$ we find $\sec{\theta} = \frac{x}{5}$ and so $x = 5\sec{\theta}$ and $dx = 5\sec{\theta}\tan{\theta}d\theta$

Then

$\int \frac{1}{\sqrt{x^2-49}}dx = 5 \int \frac{\sec{\theta}\tan{\theta}d\theta}{\tan{\theta}}$

Then after simplification:

$\int \sec{\theta}d\theta = ln|\sec{\theta} + \tan{\theta}| + C$

and solving for theta using the triangle:

$ln|\frac{x}{7} + \frac{\sqrt{x^2-49}}{7}| + C$

The autograder marked me wrong. The correct answer was:

$ln|\frac{1}{7}(\sqrt{x^2-49} + x)| + C$

Which looks to be the same to me. I cannot see the difference. Even the plots look the same. I must be forgetting a property of logs that makes these distinct. Can you help me?

 

[PeroK]

I'm assuming an autograder is some software? If so, its answer looks neater to me, but the answers are clearly equivalent.

 

[Orodruin]

Some autograders are just ... sub-par ...