Question

If $\int \frac{\cos x -\sin x +1- x }{ e ^{ x }+\sin x + x } dx =\ln (f( x ))+ g ( x )+ C$ where $C$ is the constant of integration and $f( x )$ is positive, then $f( x )+g( x )$ has the value equal to

• A. $e ^{ x }+\sin x +2 x$
• B. $e ^{ x }+\sin x$
• C. $e ^{ x }-\sin x$
• D. $e ^{ x }+\sin x + x$

Answer 1

Answer: (B)

$ I=\int \frac{\left( e ^{ x }+\cos x +1\right)-\left( e ^{ x }+\sin x + x \right)}{ e ^{ x }+\sin x + x } dx$
$=\ln \left( e ^{ x }+\sin x + x \right)- x + C$
$\therefore f ( x )= e ^{ x }+\sin x + x \text { and } g ( x )=- x$
$ f ( x )+ g ( x )= e ^{ x }+\sin x $