How do you integrate #1/(x^2+25)#?

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Answer 1 · 2018-06-27T16:20:15

#int dx/(x^2+25) = 1/5arctan(x/5) +C#

Substitute:

#x=5t#, #dx = 5dt#

to have:

#int dx/(x^2+25) = int (5dt)/((5t)^2+25)#

#int dx/(x^2+25) = 1/5 int dt/(t^2+1)#

#int dx/(x^2+25) = 1/5arctan t +C#

#int dx/(x^2+25) = 1/5arctan(x/5) +C#