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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. The `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`

A

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

B

`e^(x)+sinx`

C

`e^(x)-sinx`

D

`e^(x)+sinx+x`

Text Solution

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The correct Answer is:
B
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