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Q.

(x)3(x)5+x4dx=alogxkxk+1+c then the value of a & k  respectively are

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a

does not exist

b

23&32

c

32&2

d

32&23

answer is D.

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Detailed Solution

(x)3(x)5+x4dx

=x3/2x5/2+x4dx

=x3/2x5/21+x3/2dx =1x5/2x-3/2+1dx =1t-23dt                        Put x-3/2+1=t =-23log t+C                           -32x-5/2dx=dt =-23logx-3/2+1+C =-23log1+x3/2x3/2+C =23log x3/21+x3/2+C a=2/3, K=3/2

 

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