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Question : 74 of 160
Marks:
+1,
-0
Solution:
I=∫(‌)2‌dxLet
y=log‌x. Then,
x=ey⇒dx=ey⋅dySo,
I=∫(‌)2eydy=∫(‌)eydy⇒∫(‌−‌)eydySince,
∫(f(y)+f′(y))eydy=f(y)ey+CHere,
f(y)=‌ and
f′(y)=‌∴‌‌I‌=f(y)ey+C=‌ey+C‌=‌⋅elog‌x+C‌=‌+C
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