Question

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!

Answer 1

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`.

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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!

Answer 2

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.