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Ifint(3e^x-5e^(-1))/(4e^x+5e^(-x))dx=a x...

`Ifint(3e^x-5e^(-1))/(4e^x+5e^(-x))dx=a x+bln(4e^x+5e^(-x))+C ,t h e n` `a=-1/8,b=7/8` (b) `a=1/8,b=7/8` `a=-1/8,b=-7/8` (d) `a=1/8,b=-7/8""`

A

`a= -(1)/(8), b=(7)/(8)`

B

`a= (1)/(8), b=(7)/(8)`

C

`a= -(1)/(8), b= -(7)/(8)`

D

`a= (1)/(8), b= -(7)/(8)`

Text Solution

Verified by Experts

The correct Answer is:
A
`int(3e^(x)-5e^(-x))/(4e^(x)+5e^(-x))dx=ax+b " In " (4e^(x)+5e^(-x))+C`
Differentiating both sides, we get
`(3e^(x)-5e^(-x))/(4e^(x)+5e^(-x))=a+b ((4e^(x)-5e^(-x)))/(4e^(x)+5e^(-x))`
or `3e^(x)-5e^(-x)=a(4e^(x)+5e^(-x))=b(4e^(x)-5e^(-x))`
Comparing the coefficients of like terms on both sides, we get
`3=4(a+b),-5=5a-5b" or " a= -(1)/(8), b=(7)/(8)`
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