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

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

First, use partial fractions to break up the integrand into simpler fractions:
(1+x)^3/(x^2+1) = A(x+1) + B(x^2+1) + C
Multiplying both sides by (x^2+1) and substituting x = i and x = -i, we get:
A(i+1) + B(i^2+1) + C = 0
A(-i+1) + B((-i)^2+1) + C = 0
Simplifying and solving for A, B, and C, we get:
A = 1/2, B = 3/4, C = -1/4

Now, substitute the partial fractions back into the integral:
∫(1+x)^3/(x^2+1) dx = ∫(1/2)(x+1)/(x^2+1) dx + ∫(3/4)(x^2+1)/(x^2+1) dx - ∫(1/4) dx
Simplifying and integrating, we get:
(1/2)ln(x^2+1) + (3/4)x - (1/4)arctan(x) + C

Therefore, the final answer is (1/2)ln(x^2+1) + (3/4)x - (1/4)arctan(x) + C.