What is the integral of (e^x+2)/(e^{2x)+2e^x+5} ?

The integral of (e^x+2)/(e^{2x)+2e^x+5} is 2/5 x+3/10 arctan(1/2 (e^x+1))-1/5 ln|2e^x+e^{2x}+5|+C

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integral of (e^x+2)/(e^{2x)+2e^x+5}

\int \frac{e ^{x} + 2}{e ^{2 x} + 2 e ^{x} + 5} d x

\frac{2}{5} x + \frac{3}{10} arctan (\frac{1}{2} (e^{x} + 1)) - \frac{1}{5} ln 2 e^{x} + e^{2 x} + 5 + C

Solution steps

\int \frac{e ^{x} + 2}{e ^{2 x} + 2 e ^{x} + 5} d x

Apply u-substitution

: \int \frac{u + 2}{u ( u ^{2} + 2 u + 5 )} d u

= \int \frac{u + 2}{u ( u ^{2} + 2 u + 5 )} d u

Expand

\frac{u + 2}{u ( u ^{2} + 2 u + 5 )} : \frac{u}{u ( u ^{2} + 2 u + 5 )} + \frac{2}{u ( u ^{2} + 2 u + 5 )}

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{u}{u ( u ^{2} + 2 u + 5 )} d u + \int \frac{2}{u ( u ^{2} + 2 u + 5 )} d u

\int \frac{u}{u ( u ^{2} + 2 u + 5 )} d u = \frac{1}{2} arctan (\frac{u + 1}{2})

\int \frac{2}{u ( u ^{2} + 2 u + 5 )} d u = 2 (\frac{1}{5} ln ∣ u ∣ + \frac{1}{10} (- ln u^{2} + 2 u + 5 - arctan (\frac{u + 1}{2})))

= \frac{1}{2} arctan (\frac{u + 1}{2}) + 2 (\frac{1}{5} ln ∣ u ∣ + \frac{1}{10} (- ln u^{2} + 2 u + 5 - arctan (\frac{u + 1}{2})))

Substitute back

u = e^{x}

= \frac{1}{2} arctan (\frac{e ^{x} + 1}{2}) + 2 (\frac{1}{5} ln ∣ e^{x} ∣ + \frac{1}{10} (- ln (e^{x})^{2} + 2 e^{x} + 5 - arctan (\frac{e ^{x} + 1}{2})))

Simplify

\frac{1}{2} arctan (\frac{e ^{x} + 1}{2}) + 2 (\frac{1}{5} ln ∣ e^{x} ∣ + \frac{1}{10} (- ln (e^{x})^{2} + 2 e^{x} + 5 - arctan (\frac{e ^{x} + 1}{2}))) : \frac{2}{5} x + \frac{3}{10} arctan (\frac{1}{2} (e^{x} + 1)) - \frac{1}{5} ln 2 e^{x} + e^{2 x} + 5

= \frac{2}{5} x + \frac{3}{10} arctan (\frac{1}{2} (e^{x} + 1)) - \frac{1}{5} ln 2 e^{x} + e^{2 x} + 5

Add a constant to the solution

= \frac{2}{5} x + \frac{3}{10} arctan (\frac{1}{2} (e^{x} + 1)) - \frac{1}{5} ln 2 e^{x} + e^{2 x} + 5 + C

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What is the integral of (e^x+2)/(e^{2x)+2e^x+5} ?
The integral of (e^x+2)/(e^{2x)+2e^x+5} is 2/5 x+3/10 arctan(1/2 (e^x+1))-1/5 ln|2e^x+e^{2x}+5|+C