How do you test the improper integral #int (3x)/(x+1)^4 dx# from #[0, oo)# and evaluate if possible?
1 Answer
Mar 6, 2017
Explanation:
#int_0^oo(3x)/(x+1)^4dx#
First just treating without the bounds:
#int(3x)/(x+1)^4dx#
Let
#=int(3(u-1))/u^4du=int(3/u^3-3/u^4)du#
Rewriting and integrating:
#=int(3u^-3-3u^-4)du=(3u^-2)/(-2)-(3u^-3)/(-3)#
Since
#=(-3)/(2(x+1)^2)+1/(x+1)^3=(-3(x+1)+2)/(2(x+1)^3)=(-3x-1)/(2(x+1)^2)#
So, we have:
#int_0^oo(3x)/(x+1)^4dx=[(-3x-1)/(2(x+1)^2)]_0^oo#
In order to "evaluate" this at infinity, take the limit:
#=(lim_(xrarroo)(-3x-1)/(2(x+1)^2))-((-3(0)-1)/(2(0+1)^2))#
The degree of the numerator exceeds that of the denominator, so the limit is
#=0-((-1)/(2))#
#=1/2#