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Question : 71 of 160
Marks:
+1,
-0
Solution:
‌‌ (b) ‌∫‌(˙x+√x)‌dx‌=∫(√xe√x+e√x)‌dx=∫√xe√x‌dx+∫e√x‌dx Now,
∫√x⋅e√x‌dxLet
x=p2⇒dx=2pdp ‌=∫√xe√x‌dx=∫pep(2pdp)=2‌∫p2ep⋅dp‌=2[p2ep−2‌∫pepdp]‌=2[p2ep−2{pep−ep}]+K‌=2[p2pp−2pep+2ep]+K‌=2[xe√x−2√xe√x+2e√x]+K ‌∫e√x‌dx=2‌∫pepdp=2[pep−ep]=2[√xe√x−e√x]‌∴∫‌(x+√x)‌dx ‌=2xe√x−4√xe√x+4e√x+2√xe√x−2e√x+K‌=e√x(2x−2√x+2)+K‌∴A=2B=−2,C=2‌∴A+B+C=2
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