Question

Prove the following:

`int_1^3 dx/(x^2(x +1)) = 2/3 + log  2/3`

Answer 1

Let `I = int_1^3 dx/(x^2 (x + 1))`

Now, `1/(x^2 (x + 1)) = A/x + B/x^2 + C/(x + 1)`

∴ 1 ≡ Ax (x + 1) + B(x + 1) + Cx²                ....(i)

Putting x = 0 in (i), we get

1 = B (0 + 1)

⇒ B = 1

Putting x = -1 in (i), we get

= C (-1)²

⇒ C = 1

Comparing coefficients of x² on the sides of (i), we get

∴ 0 = A + C

∴ A = - C = - 1

⇒ A = -1

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

∴ `int_1^3 1/(x^2 (x + 1))`dx

`= - int_1^3 1/x "dx" + int_1^3 1/x^2 "dx" + int_1^3 1/(x + 1)`dx

`= [- log |x| + x^-1/-1 + log |x + 1|]_1^3`

`= [- 1/x + log |(x + 1)/x|]_1^3 = (-1/3 + 1) + log  4/3 - log 2`

`= 2/3 + log (4/3 xx 1/2)`

`= 2/3 + log  2/3`