_ -1 ^1 d d x ( ⁻¹ 1 x ) d x is

Step-by-step solution

Let f(x) = ⁻¹ ( 1 x ) . The integral is _ -1 ¹ f'(x) dx . The function f(x) is discontinuous at x = 0 . Therefore, the integral must be evaluated as the sum of two improper integrals: _ -1 ⁰ f'(x) dx + ₀¹ f'(x) dx = _ t 0⁻ [f(t) - f(-1)] + _ t 0⁺ [f(1) - f(t)] . We know that _ t 0⁻ ⁻¹ ( 1 t ) = - 2 and _ t 0⁺ ⁻¹ ( 1 t ) = 2 . Also, f(-1) = ⁻¹(-1) = - 4 and f(1) = ⁻¹(1) = 4 . Substituting these values: ( - 2 - (- 4 ) ) + ( 4 - 2 ) = ( - 2 + 4 ) + ( 4 - 2 ) = - 4 - 4 = - 2 . Answer: - 2