Question

Evaluate:

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

Answer 1

We can write equation as

`(x^2 + x + 1)/((x + 2)(x^2 + 1)) dx = A/((x + 2)) + (Bx + C)/((x^2 + 1))`

Cancelling denominator:

x² + x + 1 = A(x² + 1) + (Bx + C) (x + 2)

Putting x = −2

(−2)² + (−2) + 1 = A((−2)² + 1) + 0

4 − 2 + 1 = 5A

`3/5` = A

Putting x = 0

x² + x + 1 = A(x² + 1) + (Bx + C) (x + 2)

0 + 0 + 0 = A(0 + 1) + (0 + C) (0 + 2)

1 = A + 2C

1 = `3/5 + 2C`

`2/5` = 2C

C = `1/5`

Putting x = 1

x² + x + 1 = A(x² + 1) + (Bx + C) (x + 2)

1 + 1 + 1 = 2A + (B + C) (3)

3 = 2A + 3 (B + C)

`3 = 2(3/5) + 3 (B + 1/5)`

`3 - 6/5 = 3(B + 1/5)`

`9/5 = 3(B + 1/5)`

`3/5 - 1/5` = B

B = `2/5`

Thus, `(x^2 + x + 1)/((x + 2)(x^2 + 1)) dx = A/((x + 2)) + (Bx + C)/((x^2 + 1))`

`(x^2 + x + 1)/((x + 2)(x^2 + 1)) dx = 3/(5(x + 2)) + (1(2x + 1))/(5 (x^2 + 1))`

Hence, our equation becomes:

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

= `int 3/(5(x^2 + 1)) dx + int 1/5 (2x + 1)/(x^2 + 1) dx`

= `int 3/(5(x^2 + 1)) dx + 1/5 int (2x)/(x^2 + 1) dx + 1/5 int 1/(x^2 + 1) dx`

I₁ = `3/5int1/(x + 2) dx`

= `3/5 log |x + 2| + C_1`

I₂ = `1/5 int (2x)/(x^2 + 1) dx`

Let t = x² + 1

`(dt)/dx` = 2x

dt = 2x dx

Substituting,

= `1/5 int (dt)/t`

= `1/5 log |t| + C_2`

= `1/5 log |x^2 + 1| + C_2`

I₃ = `1/5 int 1/(x^2 + 1) dx`

= `1/5 tan^-1 (x) + C_3`

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

= `3/5 log |x + 2| + 1/5 log |x^2 + 1| + 1/5 tan^-1 (x) + C`  ......(Where C = C₁ + C₂ + C₃)