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B
x4+xx5+x+1+c
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C
5x4x5+x+1+c
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D
x5x5+x+1+c
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Solution
The correct option is Dx5x5+x+1+c I=∫x4(5+4x)(x5+x+1)2dx⇒I=∫x4(5+4x)x10(1+x−4+x−5)2dx⇒I=∫(5x4+4x5)x10(1+x−4+x−5)2dx ⇒I=∫5x−6+4x−5(1+x−4+x−5)2dx
Put (1+x−4+x−5)=t⇒−(5x−6+4x−5)dx=dt