Indefinite IntegrationHard
Question
The value of 
dx is equal to :
dx is equal to : Options
A.2e√x [√x - x + 1] + C
B.2e√x [x - 2√x + 1] + C
C.2e√x [x - √x + 1] + C
D.2e√x (x + √x + 1) + C
Solution
I = 
Put √x = t ⇒ 
dx = 2 dt
⇒ I = 2∫(t2 + t)dt = 2
= 2
= 2[et . t2 - tet + et] + C
= 2et [t2 - t + 1] + C = 2e√x [x - √x + 1] + C
Put √x = t ⇒ dx = 2 dt ⇒ I = 2∫(t2 + t)dt = 2
= 2 = 2[et . t2 - tet + et] + C= 2et [t2 - t + 1] + C = 2e√x [x - √x + 1] + C
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