find the indefinite integral

Question

integral{the cubic root of x+1/(the cubic root of x+1 plus the cubic root of 11-x)}

Answer 1

Not for the weak at heart...

Here is an OUTLINE of the solution

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the integrand can be rewritten as:

1/ [cube_root( (12/(x+1) - 1) + 1]

Begins with substitution U = -x-1, so that

du/dx = -1 ---> dx = -du

the integral becomes:

1/ [ cube_root( 12/u + 1) - 1 ]

Next by letting V be this entire denominator,

du = (-1/4)(12/u + 1)^(2/3) u^2 dv

and

(12/u+1)^(2/3) = (v+1)^2 or equivalently,

u^2 = 144/((x+1)^3-1)^2

the integral becomes:

-36 integral [ (v+1)^2 *dv / [ v((v+1)^3-1)^2 ]

temporarily ignoring the constant of -36,

this integrand simplifies to:

(v+1)^2 dv / [ v^3 (v^2+3v+3)^2 ]

which can be integrated via Partial Fraction Decomposition (PFD).

The PDF is:

a/v + b/v^2 + c/v^3 + (dx+e)/(v^2+3x+3) + (fx+g)/(v^2+3x+3)^2

for fixed number constants a,b,c,d,e,f,g

the v^2 term ends up cancelling out...

also along the way, the v^2+3x+3 term ends up being

(v+3)/(v^2+3v+3) which needs to be broken up into

2 separate integrals:

(1/2)(2v+3)/(v^2+3v+3) + 3/2/(v^2+3v+3)

the first is done by substitution w=v^2+3x+3 --> dw = 2v+3 dv

the second is done by completing the square in the denominator:

(v + (3/2))^2 + 3/4

letting w = (2v+3)/sqrt(3) --> dv = sqrt(3)/2 dw, which transforms

this second piece into

integral 1/(w^2+1) = arctan(w)