What's the integral of (1+x)^3/(1+x^2)?

The integral of (1+x)^3/(1+x^2) is (1/2)(1+x)^2 + (1/2)ln(1+x^2) + C.

To solve this integral, we can use the method of partial fractions. First, we write the integrand as:

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

Next, we integrate each term separately. The first term can be integrated using the power rule:

∫(1+x)^2 dx = (1/3)(1+x)^3 + C1

The second term can be integrated using the natural logarithm:

∫(x^2+1) dx = x^3/3 + x + C2

For the third term, we use the substitution u = 1+x^2, du/dx = 2x:

∫-2x^2/(1+x^2) dx = -∫x(du/u) = -ln|1+x^2| + C3

Putting it all together, we have:

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

Simplifying further, we get:

∫(1+x)^3/(1+x^2) dx = (1/2)(1+x)^2 + (1/2)ln(1+x^2) + C

where C = C1 + C2 + C3 is the constant of integration.