What is the integral of (e^x)/((e)^2)x+1 ?

The integral of (e^x)/((e)^2)x+1 is e^{x-2}(x-1)+x+C

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integral of (e^x)/((e)^2)x+1

integral

\frac{e ^{x}}{( e ) ^{2}} x + 1

e^{x - 2} (x - 1) + x + C

Solution steps

\int \frac{e ^{x}}{e ^{2}} x + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{e ^{x}}{e ^{2}} x d x + \int 1 d x

\int \frac{e ^{x}}{e ^{2}} x d x = e^{x - 2} (x - 1)

\int 1 d x = x

= e^{x - 2} (x - 1) + x

Add a constant to the solution

= e^{x - 2} (x - 1) + x + C

\frac{1}{( 2 - y ^{2} )}

\frac{1}{( a + x ^{2} )}

\frac{1}{( 1 + sin ( x ) )}

x^{- 1} 0.01

\frac{1}{cot ^{3} ( x )}

What is the integral of (e^x)/((e)^2)x+1 ?
The integral of (e^x)/((e)^2)x+1 is e^{x-2}(x-1)+x+C