Mathematics

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


ANSWER

$$e^{x}+\cos x+1$$


SOLUTION
$$\displaystyle I=\int { \frac { \cos { x-\sin { x+1-x }  }  }{ { e }^{ x }+\sin { x } +x }  } dx$$

$$\displaystyle =\int { \frac { { e }^{ x }+\cos { x } +1-{ e }^{ x }-\sin { x } -x }{ { e }^{ x }+\sin { x } +x }  } dx$$

$$\displaystyle =\int { \left( \frac { { e }^{ x }+\cos { x } +1 }{ { e }^{ x }+\sin { x+x }  } -1 \right) dx } $$

$$=\log { \left( { e }^{ 2 }+\sin { x+x }  \right) -x } $$
$$\displaystyle \because \frac { d }{ dx } \left( { e }^{ x }+\sin { x } +x \right) ={ e }^{ x }+\cos { x } +1$$
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Single Correct Medium Published on 17th 09, 2020
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