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

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

To solve this integral, we can use partial fractions. First, we factor the denominator as (x+i)(x-i). Then, we can write the integrand as:

(1+x)^4/[(x+i)(x-i)] = A/(x+i) + B/(x-i)

Multiplying both sides by (x+i)(x-i), we get:

(1+x)^4 = A(x-i) + B(x+i)

We can solve for A and B by substituting x=-i and x=i, respectively. This gives us:

A = (1-i)^4/2i = -1/2 + 2i
B = (1+i)^4/2i = -1/2 - 2i

Now, we can rewrite the integrand as:

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

We can then integrate each term separately using the substitution u = x+i for the first term and u = x-i for the second term. This gives us:

∫(-1/2 + 2i)/(x+i) dx = -1/2ln|x+i| + 2i∫dx/(x+i) = -1/2ln|x+i| + 2i ln|x+i| = (2i-1/2)ln|x+i| + C1

∫(-1/2 - 2i)/(x-i) dx = -1/2ln|x-i| - 2i∫dx/(x-i) = -1/2ln|x-i| - 2i ln|x-i| = (-2i-1/2)ln|x-i| + C2

Adding these two integrals together, we get:

∫(1+x)^4/[(x+i)(x-i)] dx = (2i-1/2)ln|x+i| + (-2i-1/2)ln|x-i| + C

Simplifying this expression, we get:

∫(1+x)^4/[(x+i)(x-i)] dx = (x^3