Mathematics integral problem

Please integrate step by step using the partial fractions method:

INT (((1+x)^2)/((x)(1+x^(2)))) dx

Thank you in advance.

There is no need to use substitution for this integral as the above answerer has done. Powers are usually written in descending order wherever possible.

Decompose the original integrand into partial fractions:
(x + 1)² / x(x² + 1) = (x² + 2x + 1) / x(x² + 1)
(x² + 2x + 1) / x(x² + 1) = A / x + (Bx + C) / (x² + 1)
x² + 2x + 1 = A(x² + 1) + (Bx + C)x
x² + 2x + 1 = Ax² + A + Bx² + Cx
x² + 2x + 1 = (A + B)x² + Cx + A
A = 1
C = 2
A + B = 1
B = 1 - A
B = 1 - 1
B = 0
(x² + 2x + 1) / x(x² + 1) = 1 / x + 2 / (x² + 1)

Integrate the original function term by term using this result:
∫ (x² + 2x + 1) / x(x² + 1) dx = ∫ [1 / x + 2 / (x² + 1)] dx
∫ (x² + 2x + 1) / x(x² + 1) dx = ∫ 1 / x dx + 2 ∫ 1 / (x² + 1) dx
∫ (x² + 2x + 1) / x(x² + 1) dx = ln|x| + 2tanˉ¹x + C
∫ (x² + 2x + 1) / x(x² + 1) dx = 2tanˉ¹x + ln|x| + C

let x = tanθ and use trig identities to simplify it into something integratable.
The key here is to know that (1 + (tanθ)^2) = (secθ)^2

∫ (1 + tanθ)^2/(tanθ(1 + (tanθ)^2)) dx
∫ ((tanθ)^2 + 2tanθ + 1)/(tanθ(secθ)^2) dx
∫ ((secθ)^2 + 2tanθ)/(tanθ(secθ)^2) dx
∫ 1/tanθ + 2/(secθ)^2 dx <-- cancelled stuff out
∫ cotθ + 2(cosθ)^2 dx
∫ cotθ + 2(1 + cos2θ)/2 dx <-- (cosθ)^2 = (1 + cos2θ)/2
∫ cotθ + 1 + cos2θ dx
ln|sinθ| + θ + 1/2sin2θ + C
ln|sin(arctanx)| + arctanx + 1/2sin(2arctanx) + C <-- x = tanθ so θ = arctanx

you can probably simplify that even further though

Follow this link and click "Show Steps" on the result.

http://www.wolframalpha.com/input/?i=%28…


Hope it helped!