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Question : 145 of 180
Marks:
+1,
-0
Solution:
‌‌ Let ‌I=∫e‌II ‌x⋅xI5‌dx
‌=x5⋅ex−∫ex⋅(5x4)‌dx
‌=x5ex−5iint‌II ‌x⋅x‌I ‌4‌dx
‌=x5ex−5[x4⋅ex−∫ex⋅4x3‌dx]
‌=x5ex−5x4ex+20iint‌II ‌x⋅x‌I ‌3‌dx
‌=x5ex−5x4ex+20[x3⋅ex−∫ex⋅3x2‌dx]
‌=x5ex−5x4ex+20x3ex−60‌∫e‌II ‌x⋅x‌I ‌2‌dx
‌=x5ex−5x4ex+20x3ex−60[x2⋅ex−∫ex⋅2x‌dx]
‌=x5e4−5x4ex+20x3ex−60x2ex+120‌∫ex⋅x‌dx
‌=x5ex−5x4ex+20x3ex−60x2ex+120[xex−∫ex‌dx]
‌=x5ex−5x4ex+20x3ex−60x2ex+120xex−120ex+C
‌=ex[x5−5x4+20x3−60x2+120x−120]+C
‌
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