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Question : 73 of 160
Marks:
+1,
-0
Solution:
We have,
∫x3(log‌x)2dx=x4 [A(log‌x)2+B(log‌x)+C‌log‌e]+K Now,
∫x3(log‌x)2dx =(log‌x)2‌∫x3dx−∫(‌(log‌x)2‌∫x3dx)dx =(log‌x)2⋅‌−∫(2‌log‌x)⋅‌⋅‌)‌d‌x =(log‌x)2(‌)−‌‌∫log‌x⋅x3dx =(log‌x)2(‌)−‌ [log‌x‌∫x3dx−∫(‌(log‌x‌∫x3dx)dx] =(log‌x)2(‌)−‌[log‌x(‌)−∫‌⋅‌dx] =(log‌x)2(‌)−‌[log‌x(‌)−‌⋅∫x3dx] =(log‌x)2(‌)−‌[log‌x(‌)−‌⋅‌]+K =x4[‌(log‌x)2−‌‌log‌x+‌‌log‌e]+K ∴‌‌A=‌,B=‌ and
C=‌ ∴A+B+C=‌−‌+‌=‌=‌
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