How do you find `int (x^2+2x+1)/((x+1)(x^2-2)) dx` using partial fractions?

We know that,

`color(red)((1)int(d/(dx)(f(x)))/(f(x))dx=ln|f(x)|+c`

`color(blue)((2)int1/(x^2-a^2)dx=1/(2a)ln|(x-a)/(x+a)|+c`

Here,

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

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

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

`=int[x/(x^2-2)+1/(x^2-2)]dx`

`=1/2int(2x)/(x^2-2)dx+int1/(x^2-(sqrt2)^2)dx`

Using `color(red)((1)) and color(blue)((2))`, we get

`I=1/2ln|x^2-2|+1/(2sqrt2)ln|(x-sqrt2)/(x+sqrt2)|+c`

Given: `int (x^2+2x+1)/((x+1)(x^2-2)) dx`

The numerator factors:

`int (x+1)^2/((x+1)(x^2-2)) dx`

Cancel the common factor:

`int (x+1)/(x^2-2) dx`

The denominator can be factored as the difference of two squares:

`int (x+1)/((x-sqrt2)(x+sqrt2) dx`

Write the partial fractions equation:

`(x+1)/((x-sqrt2)(x+sqrt2)) = A/(x-sqrt2)+B/(x+sqrt2)`

Multiply both sides by `(x-sqrt2)(x+sqrt2)`:

`x+1 = A(x+sqrt2)+B(x-sqrt2)`

Let `x = sqrt2`:

`sqrt2+1 = A(sqrt2+sqrt2)+B(sqrt2-sqrt2)`

`sqrt2+1 = A(2sqrt2)`

`A = (2+sqrt2)/4`

Let `x = -sqrt2`:

`-sqrt2+1 = A(-sqrt2+sqrt2)+B(-sqrt2-sqrt2)`

`-sqrt2+1 = B(-2sqrt2)`

`B = (2-sqrt2)/4`

Write in integral form:

`(2+sqrt2)/4 int 1/(x-sqrt2) dx + (2-sqrt2)/4 int 1/(x+sqrt2) dx`

The integrals become natural logarithms:

`(2+sqrt2)/4 ln(x-sqrt2) + (2-sqrt2)/4 ln(x+sqrt2)+ C`