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

The integral of (1+x)^2/(x^2+1) is ln|x^2+1|+x-2arctan(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)^2/[(x+i)(x-i)] = A/(x+i) + B/(x-i)

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

(1+x)^2 = A(x-i) + B(x+i)

Setting x = i, we get:

4i = A(0) + B(2i)
B = 2

Setting x = -i, we get:

4i = A(-2i) + B(0)
A = -2

Therefore, we can write the integrand as:

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

Now, we can integrate each term separately:

∫-2/(x+i) dx = -2ln|x+i| + C1

∫2/(x-i) dx = 2ln|x-i| + C2

Combining the two integrals and simplifying, we get:

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

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