Many students fall into a trap when solving problems similar to this coming up with an answer of #3ln(3)# do not make this mistake.
However, let us begin by noting that this integral does not follow our formal definition by which the integral is continuous on the given interval. Note that #x=1/4# results in a indefinite solution to the given equation.
If this original integral did not contain the value #1/4# then we could compute the integral. In this case the only reason we cannot is that #1/4# is on the interval #[0,1]#.
However in this case we are allow to set limits for the integral. From the limits we set it allows the user to determine if the integral approaches a value or diverges. In this case divergency meaning going off in two opposite directions towards #oo#.
Let us start by computing the integral if we were on a different interval that did not include #1/4#
#int_0^1(3)/(4x-1)#
#3ln(abs(4x-1))#
Now in this case let us define a new integral #int_0^b(3)/(4x-1)# and #int_b^1(3)/(4x-1)# by which we add the sums of these two integrals. We are allow to do this because in the case of these two integrals the sum is equal to the original integral.
Now since we experience a bad spot at #x=1/4# let us approach #x=1/4#
In this case let us define #limb->1^+/4[int_0^b(3)/(4x-1)]#
#= limb->1^+/4[3ln(abs(4b-1))-3ln(abs(4(0)-1))]#
#= limb->1^+/4[3ln(abs(4b-1))-3ln(1)]#
Let us then find #b# from the opposite direction.
#limb->1^-/4[int_b^1(3)/(4x-1)]#
#= limb->1^-/4[3ln(abs(4(1)-1))-3ln(abs(4b-1))]#
#= limb->1^-/4[3ln(3)-3ln(abs(4b-1))]#
let us now note that # limb->1^-/4[3ln(abs(4b-1))]= -oo#
Remember that # limb->0 ln(b) = -oo# in this case our interior function approaches zero. In cases like this we are allowed to substitute or shall I say formulate a new limit by which b approaches zero.
Now from summing and replacing our limits that:
#{limb->1^+/4[3ln(abs(4b-1))-3ln(1)]}+{limb->1^-/4[3ln(3)-3ln(abs(4b-1))]}#
#[3ln(3)+oo]-[oo]#
since #ln(1)=0#
We now have an indefinite form. In this case #oo-oo# meaning our original integral approaches the area of #oo# minus #oo# #therefore# one point on the graph bends towards negative infinity to make an area beneath the curve of #oo# and the other end of our function bends in the positive direction such the make another area of #oo#
