MHT CET202619 April 2026Morning ShiftMathematicsIndefinite IntegrationActual
The value of integral dx ^2 x + ^2 x is...
Options
- A-1 2 x + 1 2 2 ⁻¹ ( x 2 ) + c
- B-1 2 x - 1 2 2 ⁻¹ ( x 2 ) + c
- C1 2 x - 1 2 2 ⁻¹ ( x 2 ) + c
- D1 2 x + 1 2 2 ⁻¹ ( x 2 ) + c
Correct answer
B. -1 2 x - 1 2 2 ⁻¹ ( x 2 ) + c
Step-by-step solution
Given integral is I = dx ^2 x + ^2 x Rewriting the denominator: ^2 x + ^2 x = ^2 x + ^2 x ^2 x = ^2 x ( ^2 x + 1 ^2 x ) = ^2 x ( ^2 x + 1) Substituting this back into the integral: I = dx ^2 x ( ^2 x + 1) Multiplying the numerator and the denominator by ^2 x : I = ^2 x ^2 x ( ^2 x + 1) ^2 x dx I = ^2 x ^2 x (1 + ^2 x) dx Using the identity ^2 x = 1 + ^2 x : I = ^2 x ^2 x (2 + ^2 x) dx Let x = t , then ^2 x dx = dt . The integral becomes: I = dt t^2 (t^2 + 2) Using partial fractions: 1 t^2 (t^2 + 2) = 1 2 ( 1 t^2 -