I need to check if I am right solving this integral

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SUMMARY

The integral of the function \(\int{\frac{1}{a^2+x^2}dx}\) is definitively solved as \(\frac{1}{a}\tan^{-1}\frac{x}{a}+C\). Additionally, the integral \(\int{\left(5x^{3/5}-\frac{3}{2+x^2}\right)dx}\) evaluates to \(\frac{25}{8}x^{8/5}-\frac{3}{\sqrt{2}}\tan^{-1}\frac{x}{\sqrt{2}}+C\). The discussion confirms that the first integral's solution is accurate, addressing the user's concerns about their understanding of the material.

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TL;DR
I have a list of elementary integrals, and among them one that involves arctangent; the example I am dealing with is a combination I will propose in the next discussion paragraph.
Hi, PF

1-The elementary integral is ##\displaystyle\int{\displaystyle\frac{1}{a^2+x^2}dx}=\displaystyle\frac{1}{a}\tan^{-1}\displaystyle\frac{x}{a}+C##

2-The example is ##\displaystyle\int{\Big(5x^{3/5}-\displaystyle\frac{3}{2+x^2}\Big)dx}=\displaystyle\frac{25}{8}x^{8/5}-\displaystyle\frac{3}{\sqrt{2}}\tan^{-1}\displaystyle\frac{x}{\sqrt{2}}+C##

The question is: does the first statement agree with the solution showed?; any comment?

Greetings!

PD: I post without preview.
 
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Yes, it does. What is your uneasiness ?
 
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anuttarasammyak said:
Yes, it does. What is your uneasiness ?
Thanks a lot! I needed some help. I confess maths are among my favorite fields, but I am not specially good at them. I was quite sure, but still wanted to share with somebody. It was just some kind of necessity to put things in common; just ease my loneliness at this ground so interesting to me.
P&L.
Greetings!
 
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