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 If (x)5(x)7+x6dx=alnxkxk+1+c, the value of a and k, respectively, are

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a
52 and 25
b
25 and 52
c
52 and 2
d
None of these

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detailed solution

Correct option is B

∫dx(x)2+(x)7=∫dx(x)71+1(x)5 Put 1(x)5=y,dydx=−52(x)7∴I=∫−2dy5(1+y)=−25ln⁡|1+y|+c=25ln⁡11+1(x)5∴a=25,k=52


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