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int(cosx-sinx+1-x)/(e^(x)+sinx+x)dx=log(...

`int(cosx-sinx+1-x)/(e^(x)+sinx+x)dx=log_(e)(f(x))+g(x)+C` where C is the constant of integration and f(x) is positive. Then `f(x)+g(x)` has the value equal to

A

`e^(x)+sinx+2x`

B

`e^(x)+sinx`

C

`e^(x)-sinx`

D

`e^(x)+sinx+x`

Text Solution

Verified by Experts

The correct Answer is:
B
`I=int((e^(x)+cosx+1)-(e^(x)+sinx+x))/(e^(x)+sinx+x)`
`=log_(e)(e^(x)+sinx+x)-x+C`
`therefore" "f(x)=e^(x)+sin x+x and g(x)=-x`
`" "f(x)+g(x)=e^(x)+sinx`
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