What's the integral of (x^3+1)/(x^2+4)?

The integral of (x^3+1)/(x^2+4) is (1/2)x^2 + (1/2)ln(x^2+4) + (1/4)arctan(x/2) + C.

To solve this integral, we can use partial fraction decomposition. First, we factor the denominator as (x^2+4) = (x+2i)(x-2i). Then, we can write the integrand as:

(x^3+1)/(x^2+4) = A(x+2i) + B(x-2i) + Cx + D

Multiplying both sides by (x^2+4) and setting x = -2i, -2i, 0, and 2i, we can solve for A, B, C, and D. We get:

A = -i/4, B = i/4, C = 1/2, D = -1/4

Substituting these values back into the partial fraction decomposition, we get:

(x^3+1)/(x^2+4) = (-i/4)(x+2i) + (i/4)(x-2i) + (1/2)x - (1/4)

Now, we can integrate each term separately. The first two terms can be integrated using u-substitution:

∫(-i/4)(x+2i) dx = (-i/4)(1/2)x^2 - (i/4)(2i)x + C1 = (-1/8)x^2 - (1/8)ix + C1

∫(i/4)(x-2i) dx = (i/4)(1/2)x^2 - (i/4)(-2i)x + C2 = (1/8)x^2 + (1/8)ix + C2

The third term can be integrated using the power rule:

∫(1/2)x dx = (1/4)x^2 + C3

The fourth term can be integrated using the arctangent rule:

∫(-1/4) dx = (-1/4)x + C4

Putting it all together, we get:

∫(x^3+1)/(x^2+4) dx = (-1/8)x^2 - (1