Help with an indefinite integral?

Question

#int 1/(4+(7x+1)^2)dx#

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Answer 1 · 2018-03-05T20:05:32

#int 1/(4+(7x+1)^2)dx = 1/14tan^-1((7x+1)/2)+C#

Let #u = 7x+1#, then #du = 7dx# or #dx = 1/7du#

#int 1/(4+(7x+1)^2)dx = 1/7int 1/(u^2+4)du#

We recognize the form #1/(u^2+a^2)du = 1/atan^-1(u/a)+C#

#int 1/(4+(7x+1)^2)dx = 1/14tan^-1(u/2)+C#

Reverse the substitution for u:

#int 1/(4+(7x+1)^2)dx = 1/14tan^-1((7x+1)/2)+C#