How do you find the integral of #(x+6)/(x+10) dx#?

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Answer 1 · 2015-07-05T12:08:50

I found: #x-4ln|x+10|+c#

You can write:
#int(x+6)/(x+10)dx=int[1-4/(x+10)]dx=# integrating:
#=x-4ln|x+10|+c#

Answer 2 · 2015-07-05T16:22:06

There is a small trick to this some people don't see.

#int (x+6)/(x+10)dx#

#= int (x+6+4-4)/(x+10)dx#

#= int (x+10-4)/(x+10)dx#

#= int (x+10)/(x+10)-4/(x+10)dx#

#= int 1 - 4/(x+10)dx#

#= color(blue)(x - 4ln|x+10| + C)#