How do you test the improper integral #int (3x)/(x+1)^4 dx# from #[0, oo)# and evaluate if possible?

Question

Charles Anctil · Accepted Answer

#int_0^oo(3x)/(x+1)^4dx=1/2#

#int_0^oo(3x)/(x+1)^4dx#

First just treating without the bounds:

#int(3x)/(x+1)^4dx#

Let #u=x+1#. Thus #du=dx# and #x=u-1#:

#=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 #u=x+1#:

#=(-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#.

#=0-((-1)/(2))#

#=1/2#

Ezra Adams · Answer

undefined To test the improper integral ∫(3x)/(x+1)^4 dx from [0, ∞):

1. Check if the integral is improper due to infinity in the interval.
2. Determine if the integrand has singularities in the interval.
3. Apply the limit definition of improper integrals to evaluate the integral.

To evaluate the integral if possible:
1. Use partial fraction decomposition to rewrite the integrand.
2. Integrate each term separately.
3. Apply limits as needed.

The solution involves some algebraic manipulation, and the final result may be expressed in terms of natural logarithms and limits.