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Question : 73 of 160
Marks:
+1,
-0
Solution:
I=∫(log‌x)2x3dx Using integration by parts,
I=(log‌x)2‌∫x3dx−∫(‌∫x3dx)dx =−∫×dx =−‌∫(log‌x)x3dx =− (log‌x)‌∫x3dx−∫‌∫x3dx)‌d‌x] =−[−∫×dx] =−(log‌x)x4+×+C =[8(log‌x)2−4(log‌x)+1]+C ∴f(x)=8(log‌x)2−4(log‌x)+1
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