COMEDK2024Morning ShiftMathematicsIndefinite IntegrationActual
The value of 1 x+ x-1 d x is
Options
- A(x+ x-1 )+C
- B(x+ x-1 )+ ⁻¹ ( x-1 x )+C
- C(x-1+ x-1 )+ 1 3 | x-2- 3 x-2+ 3 |+C
- D(x+ x-1 )- 2 3 ⁻¹ ( 2 x-1 +1 3 )+C
Correct answer
D. (x+ x-1 )- 2 3 ⁻¹ ( 2 x-1 +1 3 )+C
Step-by-step solution
Let I = 1 x + x-1 dx . Substitute t^2 = x - 1 , so x = t^2 + 1 and dx = 2t dt . The integral becomes I = 2t t^2 + 1 + t dt = 2t t^2 + t + 1 dt . Rewrite the numerator as 2t + 1 - 1 : I = 2t + 1 t^2 + t + 1 dt - 1 t^2 + t + 1 dt . The first part is (t^2 + t + 1) . For the second part, complete the square in the denominator: t^2 + t + 1 = (t + 1 2 )^2 + 3 4 . Using 1 u^2 + a^2 du = 1 a ⁻¹( u a ) , we get 1 (t + 1/2)^2 + ( 3 /2)^2 dt = 2 3 ⁻¹( 2t + 1 3 ) . Thus, I = (t^2 + t + 1) - 2 3 ⁻¹( 2t + 1 3 ) + C . Substitutin