Question

How do you find the antiderivative of `int 1/(x^2(1+x^2)) dx`?

Answer 1

Partial fraction expansion gives you two trivial integrals:

`int1/(x^2(x^2 + 1))dx = int1/x^2dx - int1/(x^2 + 1)dx`

Explanation:

Use partial fraction expansion:

`1/(x^2(x^2 + 1)) = A/x + B/x^2 + (Cx +D)/(x^2 + 1)`

`1 = A(x(x^2 + 1)) + B(x^2 + 1) + (Cx +D)(x^2)`

Let `x = 0`:

B = 1

`1 - (x^2 + 1) = A(x(x^2 + 1)) + (Cx +D)(x^2)`

Let x = 1:

`-1 = A(1(1^2 + 1)) + (C + D)(1^2)`

-1 = 2A + C + D

Let x = -1

`1 - (-1^2 + 1) = A(-1(-1^2 + 1)) + (C-1 +D)(-1^2)`

-1 = -2A - C + D

D = -1

A = C = 0

Check:

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

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

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

`1/(x^2(x^2 + 1))` This checks.

`int1/(x^2(x^2 + 1))dx = int1/x^2dx - int1/(x^2 + 1)dx`

Answer 2

`-1/x-arctanx+C`

Explanation:

This method avoids partial fractions and uses a trig substitution.

`I=int1/(x^2(1+x^2))dx`

Let `x=tantheta`. This implies that `dx=sec^2thetad theta`. Thus:

`I=intsec^2theta/(tan^2theta(1+tan^2theta))d theta`

Since `1+tan^2theta=sec^2theta`:

`I=int1/tan^2thetad theta=intcot^2thetad theta`

Note that `cot^2theta=csc^2theta-1`:

`I=intcsc^2thetad theta-intd theta`

These are common integrals:

`I=-cottheta-theta+C`

Note that `tantheta=x`, so `cottheta=1/x` and `theta=arctanx`.

`I=-1/x-arctanx+C`