How to integrate (x^2+1)/(x^3+1)?

To integrate (x^2+1)/(x^3+1), use partial fractions and substitution.

First, factorise the denominator using the sum of cubes formula:
x^3+1 = (x+1)(x^2-x+1)

Then, use partial fractions to write the integrand as:
(x^2+1)/(x^3+1) = A/(x+1) + Bx+C/(x^2-x+1)

To find A, multiply both sides by (x+1) and substitute x=-1:
A = (-1^2+1)/((-1+1)(-1^2-1+1)) = 1/3

To find B and C, multiply both sides by (x^2-x+1) and substitute x=0 and x=1:
B+C = 0
B+C = 2/3

Solving these equations gives B=1/3 and C=-1/3.

Substitute these values back into the partial fractions expression:
(x^2+1)/(x^3+1) = 1/(3(x+1)) + (x+1)/(3(x^2-x+1)) - 1/(3(x^2-x+1))

Now, use the substitution u=x^2-x+1 and du=(2x-1)dx:
∫(x^2+1)/(x^3+1) dx = (1/3)ln|x+1| + (1/6)ln|x^2-x+1| - (1/3)∫du/u

The final integral can be evaluated using the natural logarithm:
∫du/u = ln|u| + C = ln|x^2-x+1| + C

Therefore, the final answer is:
∫(x^2+1)/(x^3+1) dx = (1/3)ln|x+1| + (1/6)ln|x^2-x+1| - (1/3)ln|x^2-x+1| + C
= (1/3)ln|x+1| - (1/6)ln|x^2-x+1| + C