If ∫(√x)5(√x)7+x6dx=aln(xkxk+1)+c, the values of a and k respectively are
A
52 and 25
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B
25 and 52
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C
52 and 2
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D
None of these
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Solution
The correct option is B25 and 52 I=∫dx(√x)+(√x)7=∫dx(√x)7[1+1(√x)5]
Put 1(√x)5=y⇒dydx=−52(√x)7)∴I=∫−2dy5(1+y)=−25ln|1+y|+c=25ln[(√x)5(√x)5+1]+C∴a=25,k=52