What is #int 9/(x^2sqrt(4x+1))dx#?

I found this integral question I asked in Yahoo! Answers over 4 years ago. It was answered, but I think we can do better!

2 Answers
Dec 25, 2016

I got:

#18ln|(sqrt(4x + 1) + 1)/(sqrt(4x + 1) - 1)| - (9sqrt(4x + 1))/(x) + C#

and to prove that it worked, Wolfram Alpha shows that the derivative of this is the original integrand for #x > 0#.


Let's take a crack at this. I can expect some form of substitution, and probably partial fractions.

Let:

#u = sqrt(4x + 1)#

#du = 2/(sqrt(4x + 1))dx#
#=> dx = u/2du#

#x^2 = (u^2 - 1)^2/16#

This gives:

#=> 9 int 1/(x^2sqrt(4x + 1))dx#

#= 9 int 16/((u^2 - 1)^2cancel(u))cancel(u)/2du#

#= 72 int 1/((u^2 - 1)^2)du#

Yep, partial fractions. With this, we have:

#int 1/(u^2 - 1)^2#

#= int A/(u+1) + B/(u+1)^2 + C/(u-1) + D/(u-1)^2du#

Getting common denominators allows us to cancel out the denominators and focus on the numerator:

#=> A(u + 1)(u - 1)^2 + B(u - 1)^2 + C(u - 1)(u + 1)^2 + D(u + 1)^2#

#= (Au + A)(u^2 - 2u + 1) + Bu^2 - 2Bu + B + (Cu - C)(u^2 + 2u + 1) + Du^2 + 2Du + D#

#= color(darkblue)(Au^3) - color(orange)(2Au^2) + color(cyan)(Au) + color(orange)(Au^2) - color(cyan)(2Au) + color(red)(A) + color(orange)(Bu^2) - color(cyan)(2Bu) + color(red)(B) + color(darkblue)(Cu^3) + color(orange)(2Cu^2) + color(cyan)(Cu) - color(orange)(Cu^2) - color(cyan)(2Cu) - color(red)(C) + color(orange)(Du^2) + color(cyan)(2Du) + color(red)(D)#

Collect like terms (making sure terms are grouped by addition!) to get:

#= Au^3 + Cu^3 color(red)(cancel(color(black)(- 2Au^2 + Au^2)))^(-Au^2) + Bu^2 + color(red)(cancel(color(black)(2Cu^2 - Cu^2)))^(Cu^2) + Du^2 + color(red)(cancel(color(black)(Au - 2Au)))^(-Au) - 2Bu + color(red)(cancel(color(black)(Cu - 2Cu)))^(-Cu) + 2Du + A + B - C + D#

#=> (A + C)u^3 + (-A + B + C + D)u^2 + (-A - 2B - C + 2D)u + (A + B - C + D) = 0u^3 + 0u^2 + 0u + 1#

This means we have a system of equations:

#A + 0B + C + 0D = 0#
#-A + B + C + D = 0#
#-A - 2B - C + 2D = 0#
#A + B - C + D = 1#

Or, the matrix:

#[(1,0,1,0,|,0),(-1,1,1,1,|,0),(-1,-2,-1,2,|,0),(1,1,-1,1,|,1)]#

where the first four columns imply the variable #A#, #B#, #C#, or #D#, respectively, and the last column is the righthand side of the original system of equations.

We can use elementary row operations to simplify this matrix into one that gives a more obvious answer.

Let #cR_i + R_j# denote multiplying row #i# by a constant #c#, adding the result to row #j#, and storing it in row #j#. Then:

#stackrel(R_1 + R_2" ")(->)[(1,0,1,0,|,0),(0,1,2,1,|,0),(-1,-2,-1,2,|,0),(1,1,-1,1,|,1)]#

#stackrel(R_1 + R_3" ")(->)[(1,0,1,0,|,0),(0,1,2,1,|,0),(0,-2,0,2,|,0),(1,1,-1,1,|,1)]#

#stackrel(-R_1 + R_4" ")(->)[(1,0,1,0,|,0),(0,1,2,1,|,0),(0,-2,0,2,|,0),(0,1,-2,1,|,1)]#

#stackrel(-R_2 + R_4" ")(->)[(1,0,1,0,|,0),(0,1,2,1,|,0),(0,-2,0,2,|,0),(0,0,-4,0,|,1)]#

This is far enough to find that:

#-4C = 1 => C = -1"/"4#

#-2B + 2D = 0 => B = D#

#B + 2C + D = 0 => C = -D = -B#

#A = -C#

Therefore, our solutions are:

#color(green)(A = B = D = 1"/"4)#
#color(green)(C = -1"/"4)#

That means:

#int9/(x^2sqrt(4x + 1))dx = 72int 1/(u^2 - 1)^2du#

#= 72/4int 1/(u+1) + 1/(u+1)^2 - 1/(u-1) + 1/(u-1)^2du#

#= 18[ln|u + 1| - 1/(u+1) - ln|u - 1| - 1/(u-1)]#

#= 18[ln|(u+1)/(u-1)| - (u-1)/(u^2 - 1) - (u+1)/(u^2 - 1)]#

#= 18[ln|(u+1)/(u-1)| - ((u-1)/(u^2 - 1) + (u+1)/(u^2 - 1))]#

#= 18[ln|(u+1)/(u-1)| - (2u)/(u^2 - 1)]#

#= 18ln|(u+1)/(u-1)| - (36u)/(u^2 - 1)#

Finally, substitute back in #u = sqrt(4x + 1)# to get:

#=> color(blue)(int 9/(x^2sqrt(4x + 1))dx)#

#= color(blue)(18ln|(sqrt(4x + 1) + 1)/(sqrt(4x + 1) - 1)| - (9sqrt(4x + 1))/(x) + C)#

Yep, still hard, even today!

Jan 14, 2017

If we hate partial fractions as I do, we can follow the method @truong-son-n used with the substitution #t=sqrt(4x+1)# up to the point:

#int9/(x^2sqrt(4x+1))dx=72int1/(t^2-1)^2dt#

Now apply the substitution #t=sectheta#. Thus #dt=secthetatanthetad theta#.

#=72int(secthetatantheta)/(sec^2theta-1)^2d theta#

Since #sec^2theta-1=tan^2theta#:

#=72int(secthetatantheta)/tan^4thetad theta=72intsectheta/tan^3theta=72intcos^2theta/sin^3theta=72intcot^2thetacscthetad theta#

From #cot^2theta=csc^2theta-1#:

#=72intcsc^3thetad theta-72intcscthetad theta#

Letting #I=intcsc^3thetad theta#, we will perform integration by parts letting

#{(u=csctheta" "=>" "du=-cscthetacotthetad theta),(dv=csc^2thetad theta" "=>" "v=-cottheta):}#

So

#I=-cscthetacottheta-intcscthetacot^2thetad theta#

Again using #cot^2theta=csc^2theta-1#:

#I=-cscthetacottheta-intcsc^3thetad theta+intcscthetad theta#

Using #I=csc^3theta#, that is, noticing the original integral is back into the mix, and also using the fairly well-known integral #intcscthetad theta=-lnabs(cottheta+csctheta)# we see that

#2I=-cscthetacottheta-lnabs(cottheta+csctheta)#

#intcsc^3thetad theta=-1/2cscthetacottheta-1/2lnabs(cottheta+csctheta)#

Which we will plug back into our original integral:

#=72(-1/2cscthetacottheta-1/2lnabs(cottheta+csctheta))-72intcscthetad theta#

#=-36cscthetacottheta-36lnabs(cottheta+csctheta)+72lnabs(cottheta+csctheta)#

#=-36cscthetacottheta+36lnabs(cottheta+csctheta)#

Our substitution was #t=sectheta#. This is a triangle where #t# is the hypotenuse, #1# is the adjacent side, and #sqrt(t^2-1)# is the opposite side.

Thus #csctheta=t/sqrt(t^2-1)# and #cottheta=1/sqrt(t^2-1)#.

Plugging these in gives:

#=-(36t)/(t^2-1)+36lnabs((t+1)/sqrt(t^2-1))#

With #t=sqrt(4x+1)#:

#=-(36sqrt(4x+1))/((4x+1)-1)+36lnabs((sqrt(4x+1)+1)/(sqrt((4x+1)-1)))#

#=-(9sqrt(4x+1))/x+36lnabs((sqrt(4x+1)+1)/(2sqrtx))+C#

Which, when graphically confirmed, is the same answer that Truong-Son found.