How do I solve these Integrals?

Question

integrate 1/(xSqrt[x^3-1]

integrate (x^4+2x^2-7)/(x^3-x^2+3x-3)

I need techniques more than solutions! Just point me in the right direction!

Answer 1

Try this to start you off for the first one:

∫[1/(x√(x^3 - 1)] dx

∫[1/(√(x²(x^3 - 1))] dx

∫[1/√(x^5 - x²)] dx

∫[1/(x^5 - x²)^(1/2)] dx

∫[(x^5 - x²)^(-1/2)] dx

Hope this makes sense, but I don't know where you could go from here?  I suppose you just need to simmplify it as possible to make your life easier.

As for the second one, just try and factor our any quantities to cancel eachother out and you might possibly get a smaller equation to integrate.

Answer 2

∫ {1 / [x√(x³ - 1)]} dx =

let √(x³ - 1) = u →

x³ - 1 = u² →

x³ = u² + 1 →

x = ³√(u² + 1)

dx = d[(u² + 1)^(1/3)] = {(1/3)(2u)(u² + 1)^[(1/3) - 1]}du = (2/3)u[(u² + 1)^(-2/3)]du →

dx = (2/3)[u / ³√(u² + 1)²] du

thus, substituting, you get:

∫ {1 / [x√(x³ - 1)]} dx = ∫ {1 / [³√(u² + 1)](u)}√(x³ - 1)]} (2/3)[u / ³√(u² + 1)²] du =

(2/3) ∫ {1 / [³√(u² + 1)](u)} [u / ³√(u² + 1)²] du =

(2/3) ∫ {u / [u ³√(u² + 1) ³√(u² + 1)²]} du =

simplify the integrand it into:

(2/3) ∫ [1 / ³√(u² + 1)³] du =

(2/3) ∫ [1 /(u² + 1)] du =

(2/3) arctan u + C

finally, substituting back u = √(x³ - 1), you get:

∫ {1 / [x√(x³ - 1)]} dx = (2/3) arctan[√(x³ - 1)] + C

as for the next one,

∫ [(x^4 + 2x^2 - 7) / (x^3 - x^2 + 3x - 3)] dx =

it's a rational integral in which the degree of the numerator is of a higher than the degree of the denominator, whereas it should be lower; therefore you should reduce the degree of the numerator by long division; unfortunately, on here it's quite hard to represent it, and then I'll use an alternative (though more complicated) degree-reducing method (however yielding the same result as long division)

∫ [(x^4 + 2x^2 - 7) / (x^3 - x^2 + 3x - 3)] dx =

rearrange the numerator as:

∫ {[x^4 + 2x^2 - 7 + (x^3 - x^3) + (x^2 - x^2) + (3x - 3x)] / (x^3 - x^2 + 3x - 3)} dx =

∫ [(x^4 + 2x^2 - 7 + x^3 - x^3 + x^2 - x^2 + 3x - 3x) / (x^3 - x^2 + 3x - 3)] dx =

∫ [(x^4 - x^3 + 3x^2 - 3x) + (x^3 - x^2 + 3x - 7)] / (x^3 - x^2 + 3x - 3)} dx =

break it up into:

∫ {[(x^4 - x^3 + 3x^2 - 3x)/ (x^3 - x^2 + 3x - 3)] +

[(x^3 - x^2 + 3x - 7) / (x^3 - x^2 + 3x - 3)]} dx =

break it up into:

∫ [(x^4 - x^3 + 3x^2 - 3x) / (x^3 - x^2 + 3x - 3)] dx +

∫ [(x^3 - x^2 + 3x - 7) / (x^3 - x^2 + 3x - 3)] dx =

factor out x from the numerator of the first integral:

∫ x [(x^3- x^2 + 3x - 3) / (x^3 - x^2 + 3x - 3)] dx +

∫ [(x^3 - x^2 + 3x - 7) / (x^3 - x^2 + 3x - 3)] dx =

and simplify it into:

∫ x dx + ∫ [(x^3 - x^2 + 3x - 7) / (x^3 - x^2 + 3x - 3)] dx =

similarly, rearrange the second integral as:

∫ x dx + ∫ {[x^3 - x^2 + 3x - 7 + (3 - 3)] / (x^3 - x^2 + 3x - 3)} dx =

∫ x dx + ∫ [(x^3 - x^2 + 3x - 7 + 3 - 3) / (x^3 - x^2 + 3x - 3)] dx =

∫ x dx + ∫ {[(x^3 - x^2 + 3x - 3) - 4] / (x^3 - x^2 + 3x - 3)} dx =

split it into:

∫ x dx + ∫ {[(x^3 - x^2 + 3x - 3) / (x^3 - x^2 + 3x - 3)] - [4 / (x^3 - x^2 + 3x - 3)]} dx =

∫ x dx + ∫ [(x^3 - x^2 + 3x - 3) / (x^3 - x^2 + 3x - 3)] dx - ∫ [4 / (x^3 - x^2 + 3x - 3)] dx =

simplifying,

∫ x dx + ∫ dx - ∫ [4 / (x^3 - x^2 + 3x - 3)] dx =

(1/2)x^2 + x - ∫ [4 / (x^3 - x^2 + 3x - 3)] dx (#) =

the remaining integral needs partial fraction decomposition,

thus you have to factor its denominator completely:

4 / (x^3 - x^2 + 3x - 3) =

4 / [x^2(x - 1) + 3(x - 1)] =

4 / [(x - 1)(x^2 + 3)] =

setting partial fraction decomposition:

4 / [(x - 1)(x^2 + 3)] = A/(x - 1) + (Bx + C)/(x^2 + 3) →

4 / [(x - 1)(x^2 + 3)] = [A(x^2 + 3) + (Bx + C)(x - 1)] / [(x - 1)(x^2 + 3)] →

4 / [(x - 1)(x^2 + 3)] = (Ax^2 + 3A + Bx^2 - Bx + Cx - C) / [(x - 1)(x^2 + 3)] →

equate the numerators:

4 = Ax^2 + 3A + Bx^2 - Bx + Cx - C →

4 = (A + B)x^2 + (-B + C)x + (3A - C)

yielding the system:

| A + B = 0 → B = - A → B = - 1

| - B + C = 0 → - (- A) + C = 0 → A + C = 0 → A = - C → A = 1

| 3A - C = 4 → 3(-C) - C = 4 → - 4C = 4 → C = - 1

therefore (see above):

4 / [(x - 1)(x^2 + 3)] = A/(x - 1) + (Bx + C)/(x^2 + 3) →

4 / [(x - 1)(x^2 + 3)] = 1/(x - 1) + (- x - 1)/(x^2 + 3) →

thus, going back to (#) above:

(1/2)x^2 + x - ∫ [4 / (x^3 - x^2 + 3x - 3)] dx =

(1/2)x^2 + x - ∫ {[1/(x - 1)] + [(- x - 1)/(x^2 + 3)]} dx =

(1/2)x^2 + x - ∫ [1/(x - 1)] dx - ∫ [(- x - 1)/(x^2 + 3)] dx =

(1/2)x^2 + x - ln | x - 1 | + ∫ [(x +1)/(x^2 + 3)] dx =

as for the remaining integral, being the denominator unfactorable, let's

attempt to change the numerator into the derivative of the denominator:

divide and multiply by 2

(1/2)x^2 + x - ln | x - 1 | + (1/2) ∫ [2(x + 1)/(x^2 + 3)] dx =

(1/2)x^2 + x - ln | x - 1 | + (1/2) ∫ [(2x + 2)/(x^2 + 3)] dx =

break it up into:

(1/2)x^2 + x - ln | x - 1 | + (1/2) ∫ {[2x /(x^2 + 3)] + [2 /(x^2 + 3)]} dx =

(1/2)x^2 + x - ln | x - 1 | + (1/2) ∫ [2x /(x^2 + 3)] dx + (1/2) ∫ [2 /(x^2 + 3)]dx =

(1/2)x^2 + x - ln | x - 1 | + (1/2) ∫ [d(x^2 + 3)] /(x^2 + 3) + ∫ [1 /(x^2 + 3)] dx =

(1/2)x^2 + x - ln | x - 1 | + (1/2) ln (x^2 + 3) + ∫ [1 /(x^2 + 3)] dx =

as for the remaining integral, factor out 3 from the denominator:

(1/2)x^2 + x - ln | x - 1 | + (1/2) ln (x^2 + 3) + ∫ dx /{3 [(x^2)/3)] + 1} =

(1/2)x^2 + x - ln | x - 1 | + (1/2) ln (x^2 + 3) + (1/3) ∫ dx /{[(x^2)/3)] + 1} =

rewrite the square as:

(1/2)x^2 + x - ln | x - 1 | + (1/2) ln (x^2 + 3) + (1/3) ∫ dx /[(x/√3)^2 + 1] =

finally, in order to turn the numerator into he derivative of (x/√3),

multiply and divide by √3:

(1/2)x^2 + x - ln | x - 1 | + (1/2) ln (x^2 + 3) + (1/3)√3 ∫ (1/√3)dx /[(x/√3)^2 + 1] =

(1/2)x^2 + x - ln | x - 1 | + (1/2) ln (x^2 + 3) + (1/√3) ∫ [d(x/√3)] /[(x/√3)]^2 + 1] =

(1/2)x^2 + x - ln | x - 1 | + (1/2) ln (x^2 + 3) + (1/√3) arctan(x/√3) + C

in conclusion:

∫ [(x^4 + 2x^2 - 7) / (x^3 - x^2 + 3x - 3)] dx =

(1/2)x^2 + x - ln | x - 1 | + (1/2) ln(x^2 + 3) + (1/√3) arctan(x/√3) + C

I hope it helps...

Bye!

Answer 3

Remember the power rule. That might be enough to get you through it.