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Question

If ∫3ex−5e−x4ex+5e−xdx=Ax+Bln|4ex+5e−x|+C, then which of the following is/are true (where A,B are fixed constants and C is constant of integration)

A
A=18
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B
A=18
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C
B=78
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D
B=78
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Solution

The correct option is C B=78∫3ex−5e−x4ex+5e−xdx=Ax+Bln(4ex+5e−x)+C Differentiating w.r.t. x on both sides, we get 3ex−5e−x4ex+5e−x=A+B(4ex−5e−x)(4ex+5e−x) ⇒3ex−5e−x=A(4ex+5e−x)+B(4ex−5e−x) Comparing the coefficients of terms containing ex and e−x on both sides, we get 4(A+B)=3⇒A+B=34 5A−5B=−5⇒A−B=−1 Solving these equations, we get A=−18,B=78 ​​​​​​

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