.
#intln(1+4x^2)/(1+4x^2)dx#
Let's factor the denominator:
#intln(1+4x^2)/((2x-i)(2x+i))dx#
Now, we can do partial fraction expansion and separate the argument of the integral into two fractions:
#ln(4x^2+1)/((2x-i)(2x+i))=(Aln(4x^2+1))/(2x-i)+(Bln(4x^2+1))/(2x+i)=#
#(A(2x+i)ln(4x^2+1)+B(2x-i)ln(4x^2+1))/((2x-i)(2x+i))=#
#(2Axln(4x^2+1)+Ailn(4x^2+1)+2Bxln(4x^2+1)-Biln(4x^2+1))/((2x-i)(2x+i))=#
#((2A+2B)xln(4x^2+1)+(Ai-Bi)ln(4x^2+1))/((2x-i)(2x+i))#
Comparing this with the argument of the integral, we find:
#2A+2B=0, :. A=-B or B=-A#
#Ai-Bi=1, :. Ai+Ai=1, :. 2Ai=1, :.#
#A=1/(2i) and B=-1/(2i)#
Multiplying both by #(i/i)#, we get:
#A=-i/2# and #B=i/2#
Substituting these into the partial fraction, we get:
#ln(4x^2+1)/((2x-i)(2x+i))=(-iln(4x^2+1))/(2(2x-i))+(iln(4x^2+1))/(2(2x+i))#
Therefore, our integral turns into:
#intln(1+4x^2)/(1+4x^2)dx=int(iln(4x^2+1))/(2(2x+i))dx-int(iln(4x^2+1))/(2(2x-i))dx=#
We can factor out the constants and make them coefficients of integrals:
#i/2int(ln(4x^2+1))/(2x+i)dx-i/2int(ln(4x^2+1))/(2x-i)dx=i/2I-i/2II#
#I=int(ln(4x^2+1))/(2x+i)dx#
Let #u=2x+i, :. du=2dx, :. dx=1/2du#
#x=(u-i)/2, :. x^2=(u-i)^2/4#
Let's substitute:
#I=intln((u-i)^2+1)/u(1/2du)=1/2intln((u-i)^2+1)/udu#
#I=1/2intln(u^2_2ui-1+1)/udu=1/2intln(u(u-2i))/udu#
#I=1/2int(lnu+ln(u-2i))/udu#
#I=1/2intlnu/udu+1/2intln(u-2i)/udu=1/2III+1/2IV#
#III=intlnu/udu#
Let #k=lnu, :. dk=1/udu, :. du=udk#
Let's substitutte:
#III=intkdk=k^2/2#
Let's substitute back:
#III=(lnu)^2/2#
#IV=intln(u-2i)/udu#
Let's factor out #(-2i)# from the expression of the natural logarithm:
#IV=intln((-2i)(u/(-2i)+1))/udu#
#IV=intln(-2i)1/udu+intln(u/(-2i)+1)/udu#
#u/(-2i)=u/(-2i)*i/i=(iu)/(-2i^2)=(iu)/(-2(-1))=(iu)/2#
Let's plug this in:
#IV=ln(-2i)int(du)/u+intln((iu)/2+1)/udu#
#IV=ln(-2i)lnu+intln((iu)/2+1)/udu#
Let #p=(-iu)/2, :. dp=-i/2du, :. du=-2/idp=-2/i(i/i)dp#
#du=(-2i)/i^2dp=(-2i)/(-1)dp=2idp#
#p=(-iu)/2, :. 2p=-iu, :. u=(2p)/(-i)=(2p)/(-i)*(i/i)=2ip#
Let's substitute:
#IV=ln(-2i)lnu+intln(1-p)/(2ip)(2idp)#
#IV=ln(-2i)lnu+intln(1-p)/(cancelcolor(red)(2i)p)(cancelcolor(red)(2i)dp)#
#IV=ln(-2i)lnu+intln(1-p)/pdp#
Let's factor #(-1)# out of the integral:
#IV=ln(-2i)lnu-int-ln(1-p)/pdp#
The integral is now a special integral called (Dilogarithmic Integral):
#int-ln(1-p)/pdp=Li_2(p)# and substituting back, we get:
#int-ln(1-p)/pdp=Li_2(p)=Li_2((-iu)/2)#
Let's plug this in:
#IV=ln(-2i)lnu-Li_2((-iu)/2)#
Let's plug in #III# and #IV#:
#I=1/2III+1/2IV#
#I=1/2(lnu)^2/2+1/2ln(-2i)lnu-1/2Li_2((-iu)/2)#
#I=(lnu)^2/4+1/2ln(-2i)lnu-1/2Li_2((-iu)/2)#
Let's substitute back #2x+i# for #u#:
#I=(ln(2x+i))^2/4+1/2ln(-2i)ln(2x+i)-1/2Li_2((-i(2x+i))/2)#
#II# can be solved using the same method as we used for #I#. The result will be:
#II=(ln(2x-i))^2/4+1/2ln(2i)ln(2x-i)-1/2Li_2((i(2x-i))/2)#
#intln(1+4x^2)/(1+4x^2)dx=i/2I-i/2II#
#intln(1+4x^2)/(1+4x^2)dx=i(ln(2x+i))^2/8+i/4ln(-2i)ln(2x+i)-i/4Li_2((-i(2x+i))/2)+i(ln(2x-i))^2/8+i/4ln(2i)ln(2x-i)-i/4Li_2((i(2x-i))/2)#
We can incorporate the absolute value function into the expressions of the logarithms to extend the integral's domain:
#intln(1+4x^2)/(1+4x^2)dx=i(ln(abs(2x+i)))^2/8+i/4ln(-2i)ln(abs(2x+i))-i/4Li_2((-i(2x+i))/2)+i(ln(abs(2x-i)))^2/8+i/4ln(2i)ln(abs(2x-i))-i/4Li_2((i(2x-i))/2)+C#
