Question

solve question 2 in the attachment​

Answer 1

Solution:

Given:

$\int \: \frac{ \cos \: x - \sin \: x + 1 - x}{ {e}^{x} + \sin \: x + x } dx = log(f(x)) + g(x) + c$

where C is constant.

To find :

• F[x] and • g[x]

Answer :

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

In numerator , add e^x and minus e^x , we get:

$\implies \: \int \: \frac{ \cos \: x - \sin \: x + 1 - x + {e}^{x} - {e}^{x} }{ {e}^{x} + \sin \: x + x } dx \\ \\ \implies \int \: \frac{( {e}^{x} + \cos \: x \: + 1) - ( {e}^{x} + \sin \: x + x) }{ {e}^{x} + \sin \: x + x} dx \\ \\ \implies \int \: \frac{ {e}^{x} + \cos \: x + 1}{ {e}^{x} + \sin \: x + x } dx - \int \: \cancel{ \frac{ {e}^{x} + \sin \: x + x }{ {e}^{x} \: + \sin \: x + x } }dx \\ \\ \implies \int \frac{ {e}^{x} + \cos \: x + 1 }{ {e}^{x} + \sin \: x + x } dx - \int \: dx$

•Here , we use substitution method for 1st integral:

=> Let e^x + sin x+ x= t

Differentiate both sides , w.r.t x

=> (e^x+cos x +1 ) dx = dt .

$\implies \: \int \: \frac{dt}{t} - \int \: dx \\ \\ \implies \: \: log(t) - x + c \\ \\ \implies \: log( {e}^{x} + \sin \: x \: \: + x) - x + c$

By comparing with :

Log (f(x))+g(x)+c

we get ,

• f(x) = e^x+ sin x+ x

• g(x) = -x

Answer 2

Step-by-step explanation:

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