#\int1/(x^2+a^2)dx#?

"Let #a# be a constant, and evaluate the integral."

1 Answer
F. Javier B.
Apr 13, 2018

See process below

Explanation:

#int1/(x^2+a^2)dx#

Lets do the change #x=atantheta#. With this change we have

#dx=asec^2theta d theta#. Put in the integral

#int(cancelasec^2theta d theta)/(a^cancel2(tan^2theta+1))#

But #tan^2theta+1=sec^2theta#. The we have

#int(cancelacancelsec^2theta d theta)/(a^cancel2cancelsec^2theta)=1/aint d theta=1/atheta +C#

Undoing the change #theta= arctan(x/a)#. We have finally

#int1/(x^2+a^2)dx=1/a arctan(x/a)+C#

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