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How do you find the integral of #int 17/(16+x^2)dx#?

1 Answer

Roy E. · mason m

Dec 9, 2016

#17/4 arctan(x/4)+C#. Standard form: #int1/(a^2+x^2)dx=(1/a)arctan (x/a)+C#

Explanation:

Alternatively substitute #x=4tan u# and use #1+tan^2u=sec^2u# giving the integral as #int (17 xx 4)sec^2 u/(16 sec^2 u)du# and hence the answer.

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