Question

∫[1 + x – (1/x)]e^[x+(1/x)]dx = __________ + c 

(a) (x + 1) e^[x+(1/x)] 

(b) (x – 1) e^[x+(1/x)] 

(c) – x e^[x+(1/x)] 

(d) xe^[x+(1/x)]

Answer 1

The correct option (d) xe^[x+(1/x)]   

Explanation:

I = ∫[1 + x – (1/x)]e^[x+(1/x)]dx 

= ∫e^[x+(1/x)]dx + ∫x[1 – (1/x²)]e^[x+(1/x)] ∙dx 

= I₁ + I₂ 

For I₂, Let u = x, V = e^[x+(1/x)][1 – (1/x²)] 

∴ I2 = x∫e^[x+(1/x)][1 – (1/x²)] – ∫∫e^[x+(1/x)][1 – (1/x²)]dx 

Let [x + (1/x)] = t hence [1 – (1/x²)]dx = dt 

∴ ∫e^[x+(1/x)] [1 – (1/x²)]dx 

= ∫e^t ∙ dt = e^t = e^[x+(1/x)] 

∴ I₂ = x ∙ e^[x+(1/x)] − ∫e^[x+(1/x)] dx + c 

∴ I = I₁ + I₂ 

= ∫e^[x+(1/x)]dx + x ∙ e^[x+(1/x)] – ∫e^[x+(1/x)] dx + c 

= x ∙ e^[x+(1/x)] + c