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Question : 158 of 180
Marks:
+1,
-0
Solution:
Let I=∫√xe√xdx
Put √x=t⇒x=t2⇒dx=2tdt
∴I=2‌∫tI2eIItdt
=2[t2et−∫et.2tdt]
=2[t2et−2‌∫tetdt]
=2[t2et−2(tet−∫et.1dt)]
=2[t2et−2tet+2et]+c
=(2t2−4t+4)et+c
=(2x−4√x+4)e√x+c
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