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

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

First, use partial fractions to break up the integrand into simpler fractions:
(x+1)^2/(x^2+1) = A(x+1)/(x^2+1) + B(x+1)^2/(x^2+1)^2

To find A and B, multiply both sides by the denominator and simplify:
(x+1)^2 = A(x^2+1) + B(x+1)^2
Expanding and equating coefficients, we get:
A+B = 1
2B = 1
Therefore, A = 1/2 and B = 1/2.

Substitute u = x^2+1 and du = 2x dx to simplify the integral:
∫(x+1)^2/(x^2+1) dx = ∫(1/2)(x+1)/(x^2+1) dx + ∫(1/2)(x+1)^2/(x^2+1)^2 dx
= (1/2)ln(x^2+1) + ∫(1/2)(1+2x^2+2x)/(x^2+1)^2 dx
= (1/2)ln(x^2+1) + ∫(1/2)(1/(x^2+1)^2) dx + ∫(x/(x^2+1)^2) dx + ∫(x/(x^2+1)^2) dx
= (1/2)ln(x^2+1) + (1/2)arctan(x) - (1/2)(1/(x^2+1)) + (1/2)(1/(x^2+1)) + C
= (1/2)ln(x^2+1) + (1/2)arctan(x) + C

Therefore, the integral of (x+1)^2/(x^2+1) is (1/2)ln(x^2+1) + (1/2)arctan(x) + C.