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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{2} - 2 u + 5} d u

= \int \frac{1}{u ^{2} - 2 u + 5} d u

Complete the square

u^{2} - 2 u + 5 : (u - 1)^{2} + 4

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

Apply u-substitution

: \int \frac{1}{v ^{2} + 4} d v

= \int \frac{1}{v ^{2} + 4} d v

Apply Trigonometric Substitution

: \int sec (w) d w

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

Simplify

ln tan (arctan (\frac{1}{2} (e^{x} - 1))) + sec (arctan (\frac{1}{2} (e^{x} - 1))) : ln (\frac{1}{2} e^{x} - 1 + - 2 e^{x} + e^{2 x} + 5)

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

Add a constant to the solution

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

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