COMEDK202510 May 2025Evening ShiftMathematicsIndefinite IntegrationActual
e^ ⁻¹ x (1+x^2 ) (1+x+x^2 ) d x=
Options
- Ax e^ ⁻¹ x (1+x^2 ) +c
- Be^ ⁻¹ x +c
- Cx e^ ⁻¹ x +c
- De^ ⁻¹ x (1+x^2 ) +c
Correct answer
C. x e^ ⁻¹ x +c
Step-by-step solution
Let I = e^ ⁻¹ x 1+x^2 (1+x+x^2) dx . Substitute t = ⁻¹ x , so dt = 1 1+x^2 dx . Then x = t . The integral becomes I = e^t (1 + t + ^2 t) dt . Using the identity 1 + ^2 t = ^2 t , we have I = e^t ( ^2 t + t) dt . Recall the standard integral form e^t (f(t) + f'(t)) dt = e^t f(t) + c . Here, let f(t) = t , then f'(t) = ^2 t . Thus, I = e^t t + c . Substituting back t = ⁻¹ x and t = x , we get I = e^ ⁻¹ x x + c . Answer: x e^ ⁻¹ x +c