UPSC NDA Math Model Paper 6

Let I=∫√xe√xdx
Substituting √x=t, we get:

1

2√x

dx=dt
⇒dx=2tdt
∴I=∫t×et×2tdt
Integrating by parts, taking t2 as the first function and et as the second function, we get:
⇒I=2[t2etdt−∫(

d

dt

t2‌∫etdt)dt]+C
⇒I=2t2et−4‌∫tetdt +C
Integrating ∫tetdt by parts, we get:
⇒I=2t2et−4[tetdt−∫(

d

dt

tetdt)dt] +C
⇒I=2t2et−4(tet−et)+C
⇒I=(2t2−4t+4)et+C
Back substituting √x=t, we get:
⇒I=(2x−4√x+4)e√x+C