evaluate this integral?

Question

Express the integrand as a sum of partial fractions and evaluate the integral of [ (x^3/x^2+x+1) dx ]

answer choices:

a) 48 ln(x -16) + (48/x+4) - [64/(x+4)^2] + C

b) (x^2/2) - x - 3 ln(x+1) + [1/x+1)^2] + C

c) (x^2/2) - x + 3 ln(x+1) + (1/x+1) + C

d) (x^2/2) - x + 3 ln(x+1) - (1/x+1) + C

i would pay you lol if you can figure this problem out and help understand how to solve this kind of problem, because I have no idea how to fully intergrate this problem after performing the partial fraction

i would pay you lol if you can figure this problem out and help understand how to solve this kind of problem, because I have no idea how to fully intergrate this problem after performing the partial fraction

i even asked my teacher and he said it fine like that, he didn't tell how to do it but the denominator is correctly inputed as "x" and not "2x"

Answer 1

You can't use partial fractions right away!
Why not?
Well, the degree of the numerator is bigger
than the denominator, so you must do
a long division first. That's the general rule.
To use partial fractions, the degree of the numerator
must be less than the degree of the denominator.
So, after long division, we get ∫(x^3 dx /(x^2+x+1) =
∫ (x-1) dx + ∫ dx/(x²+x+1)
= x²/2 -x + ∫ dx/(x²+x+1)
So now we have do evaluate the second integral.
Since the denominator is an irreducible quadratic
over the reals(It has no real roots), again no
partial fractions are needed.
Instead you must complete the square in the denominator
We get
∫ dx/(x²+x+1) = ∫ dx/ (x²+x+1/4 + 3/4)=
∫ dx/ [(x+1/2)² + (√3/2)²] = 2/√3*arctan( 2/√3*(x+1/2)).
Final result:
x²/2 -x + 2/√3*arctan( 2/√3*(x+1/2)) + C
So none of your choices are right.
The moral here is don't rush blindly into using
a particular method. Try to see which one(if any)
might work for you.

Answer 2

∫ [x^3/(x^2 + x + 1) ]dx

divide x^3 by (x^2 + x + 1) using synthetic division

x^2 + x + 1)x^3(x
...................x^3 +x^2+x
___________________
.......................-x^2 - x

quotient is x and remainder is -x^2 - x

so x^3/x^2 +x + 1 = x - (x^2+x)/(x^2 + x + 1)

=>x - (x^2 + x + 1-1)/(x^2 + x + 1)

=> x - 1 + 1/(x^2+x+1)

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

∫ [x^3/(x^2 + x + 1) ]dx =

∫x dx - ∫dx + ∫dx/[(x+1/2)^2 + 3/4]

for the last part put x+ 1/2 = sqrt(3)/2 tan u

dx = sqrt(3)/2 sec^2 u du

(x+1/2)^2 +3/ 4 = 3/tan^2 u + 3/4 = 3/4 sec^2u

substituting these in the integral

∫x dx - ∫dx + ∫sqrt(3)/2 sec^2 u/ 3/4 sec^2 u du

=>∫x dx - ∫dx +(2/sqrt(3)) ∫ du

=>x^2/2 - x + (2/sqrt(3) u + c

substituting back u = 2/sqrt(3)(tan^-1(x+1/2)

(1/2)x^2- x + (4/3) tan^-1(x+1/2) + c

Answer 3

None of those choices are correct.

You probably mis-copied the problem.
Check the denominator. It probably should be
x^2 + 2x + 1