What is the integral of (e^x)/(sqrt(e^{2x)-4e^x+6)} ?

The integral of (e^x)/(sqrt(e^{2x)-4e^x+6)} is ln| 1/(sqrt(2))(e^x-2)+sqrt(1/2 (-4e^x+e^{2x)+6)}|+C

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integral of (e^x)/(sqrt(e^{2x)-4e^x+6)}

\int \frac{e ^{x}}{e ^{2 x} - 4 e ^{x} + 6} d x

ln \frac{1}{2} (e^{x} - 2) + \frac{1}{2} (- 4 e^{x} + e^{2 x} + 6) + C

Solution steps

\int \frac{e ^{x}}{e ^{2 x} - 4 e ^{x} + 6} d x

Apply u-substitution

= \int \frac{1}{u ^{2} - 4 u + 6} d u

Complete the square

u^{2} - 4 u + 6 : (u - 2)^{2} + 2

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

Apply u-substitution

= \int \frac{1}{v ^{2} + 2} d v

Apply Trigonometric Substitution

= \int sec (w) d w

Use the common integral:

\int sec (w) d w = ln ∣ tan (w) + sec (w) ∣

= ln ∣ tan (w) + sec (w) ∣

Substitute back

= ln tan (arctan (\frac{1}{2} (e^{x} - 2))) + sec (arctan (\frac{1}{2} (e^{x} - 2)))

Simplify

ln tan (arctan (\frac{1}{2} (e^{x} - 2))) + sec (arctan (\frac{1}{2} (e^{x} - 2))) : ln \frac{1}{2} (e^{x} - 2) + \frac{1}{2} (- 4 e^{x} + e^{2 x} + 6)

= ln \frac{1}{2} (e^{x} - 2) + \frac{1}{2} (- 4 e^{x} + e^{2 x} + 6)

Add a constant to the solution

= ln \frac{1}{2} (e^{x} - 2) + \frac{1}{2} (- 4 e^{x} + e^{2 x} + 6) + C

What is the integral of (e^x)/(sqrt(e^{2x)-4e^x+6)} ?
The integral of (e^x)/(sqrt(e^{2x)-4e^x+6)} is ln| 1/(sqrt(2))(e^x-2)+sqrt(1/2 (-4e^x+e^{2x)+6)}|+C