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Question

Suppose A=dxx2+6x+25 and B=dxx26x27. If 12(A+B)=λ.tan1(x+34)+μ.lnx9x+3+C, then the value of (λ+μ) is .

A
1
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B
2
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C
3
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D
4
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Solution

The correct option is D 4
Given 12(A+B)=λ.tan1(x+34)+μ.lnx9x+3+C .....(1)
Given A=dxx2+6x+25
A=dx(x+3)2+42
Put x+3=t
dx=dt
A=dtt2+42
A=14tan1t4+c1
A=14tan1x+34+c1
Given B=dxx26x27
B=dx(x3)262
Put x3=u
dx=du
B=duu262
B=112ln|t6t+6|+c2
B=112ln|x9x+3|+c2
Now, 12(A+B)=12[14tan1x+34+12.6lnx9x+3]=3tan1(x+34)+lnx9x+3
Put this value in (1), we get
λ=3, μ=1λ+μ=4

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