What is the integral of (e^x)/(e^{2x)-6e^x+13} ?

The integral of (e^x)/(e^{2x)-6e^x+13} is 1/2 arctan((e^x-3)/2)+C

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integral of (e^x)/(e^{2x)-6e^x+13}

\int \frac{e ^{x}}{e ^{2 x} - 6 e ^{x} + 13} d x

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

Solution steps

\int \frac{e ^{x}}{e ^{2 x} - 6 e ^{x} + 13} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 6 u + 13} d u

= \int \frac{1}{u ^{2} - 6 u + 13} d u

Complete the square

u^{2} - 6 u + 13 : (u - 3)^{2} + 4

= \int \frac{1}{( u - 3 ) ^{2} + 4} d u

Apply u-substitution

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

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

Apply Integral Substitution

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

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

= \frac{1}{2} arctan (w)

Substitute back

= \frac{1}{2} arctan (\frac{e ^{x} - 3}{2})

Add a constant to the solution

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

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What is the integral of (e^x)/(e^{2x)-6e^x+13} ?
The integral of (e^x)/(e^{2x)-6e^x+13} is 1/2 arctan((e^x-3)/2)+C