Extremely difficult integral?

9∫dx/(x^2√(4x + 1)) = (*)

√(4x + 1) = u

4x + 1 = u^2

x = (u^2 - 1)/4

dx = u du/2

(*) = 9∫(u du)/(2u(u^2 - 1)^2/16 =

= 72∫du/(u^2 - 1)^2 = 72∫du/((u+1)^2(u-1)^2) =

= 18 ( ∫du/(u - 1)^2 - ∫du/(u - 1) + ∫du/(u + 1)^2 + ∫du/(u + 1) ) =

= 18 ( 1/(1 - u) - LN(u - 1) - 1/(1 + u) + LN(u + 1) ) + C =

= 18 ( 2u/(1 - u^2) + LN ( (u + 1)/(u - 1) ) ) + C =

= 36u/(1 - u^2) + 18LN ( (u^2 + 2u + 1)/(u^2 - 1)) + C =

= 36√(4x + 1)/(1 - (4x + 1)) + 18LN( ( 2√(4x + 1) + 4x + 2)/(4x + 1 - 1) ) + C =

= 18LN( ( √(4x + 1) + 2x + 1)/(2x) ) - 9√(4x + 1)/x + C =

Direct substitution should've worked for you.

w = √(4x +1) ⇒ x = (w² -1)/4

dw = 2/√(4x +1) dx ⇒ dx = (w/2) dw

9∫ [w/2]/[w((w² -1)/4)²] dw

72∫ dw/(w² -1)²

Now use partial fractions:

1/(w² -1)² = 1/(w +1)²(w -1)² ⇒

1/(w +1)²(w -1)² = A/(w +1)² +B/(w +1) +C/(w -1)² +D/(w -1)

1 = A(w -1)² +B(w +1)(w -1)² +C(w +1)² +D(w +1)²(w -1)

If you expand and regroup the right hand side you get:

1 = (B +D)w³ +(A -B +C +D)w² +(-2A -B +2C -D)w +A +B +C -D

If we consider each side as a function of w, then the coefficients of each term must be the same on each side of the equation, which allows us to construct a system of four equations:

B +D = 0

A -B +C +D = 0

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

A +B +C -D = 1

You should've learned in algebra how to solve this by elimination. If you've had linear algebra, you know that using elementary row operations on the augmented coefficient matrix achieves the same result, and you can do that on your calculator, which I did to get:

A = 1/4

B = 1/4

C = 1/4

D = -1/4

Whew! Now, since we determined through partial fraction decomposition that our last integrand was equal to

A/(w +1)² +B/(w +1) +C/(w -1)² +D/(w -1),

we can substitute the values from the system we just solved into that expression and integrate each term by inspection, or with a simple linear substitution:

72∫ dw/(w² -1)² = 72[A∫ dw/(w +1)² +B∫ dw/(w +1) +C∫ dw/(w -1)² +D∫ dw/(w -1)] =

72[-A/(w +1) +Bln|w +1| -C/(w -1) +Dln|w -1|] +K, where K is an arbitrary constant.

Now for simplification and back substitution. Since A, B, C, and D are multiples of 1/4, let's start by factoring out 1/4, to get rid of those pesky letters:

18[-1/(w +1) +ln|w +1| -1/(w -1) -ln|w -1|] +K

Now continuing to simplify:

18[-2w/(w² -1) +ln(|w +1|/|w -1|)] +K

18[-(√(4x +1))/x +ln(|1 +√(4x +1)|/|-1 +√(4x +1)|)] +K

∫ 9/(x²√(4x +1)) dx = -(9√(4x +1))/x +ln[(|1 +√(4x +1)|/|-1 +√(4x +1)|)^18] +K

try u² = 4 x + 1 to get to : 8 du / [ u² - 1 ]² ; now do an easy partial fraction..

I got { - 2 / [ u - 1 ] + 2 / [ u - 1 ]² + 2 / [ u + 1 ] + 2 / [ u - 1 ]² } , an easy integration

You should check my work however

permit e^x = u e^x dx = du ? e^x/[(e^x-2)(e^2x+a million)] dx = ? du / (u-2)(u^2+a million) To combine ? du / (u-2)(u^2+a million), use partial fractions a million/(u-2)(u^2+a million) = A/(u-2) + (Bu+C) /(u^2+a million) a million= A(u^2+a million) +(Bu+C)(u-2) a million= Au^2+A + Bu^2-2Bu +Cu -2C Equate the coefficients of u^2 on the two factors 0 = A+B ----(a million) Equate the coefficients of u on the two factors 0 = -2B+C ----(2) Equate the constants on the two factors a million = A-2C ----(3) remedy for A,B, and C from (a million),(2), and (3) A= 0.2 = a million/5 B= -0.2 = -a million/5 C= 0.4 = 2/5 a million/(u-2)(u^2+a million) = A/(u-2) + (Bu+C) /(u^2+a million) = a million/5(u-2) - u/5(u^2+a million) +2/5(u^2+a million) ? [du/5(u-2) = (a million/5) ln(u-2) ----(a million) -? u/5(u^2+a million) = permit u^2+a million = t 2u du=dt u du=(a million/2) dt -? u/5(u^2+a million) =- (a million/10) ln(t) = (-a million/10) ln(u^2+a million) ---(2) +2/5(u^2+a million)] = (-2/5) arctan(u) ---(3) (a million)+(2)+(3) with u replaced with e^x is: (a million/5) ln(e^x-2) -(a million/10) ln(e^2x +a million) - (2/5) arctan(e^x) + C

-((9 Sqrt[1 + 4 x])/x) + 36 ArcTanh[Sqrt[1 + 4 x]] + C

The answer is what it is.... all you can do is turn the hyperbolic functions into exponentials

Tanh [x]= Sinh[x]/Cosh[x] = (e^x - e^-x)/(e^x + e^-x)