What is the integral of (e^x)/((e^x-6)(e^{2x)+1)} ?

The integral of (e^x)/((e^x-6)(e^{2x)+1)} is 1/37 ln|e^x-6|+1/37 (-1/2 ln|e^{2x}+1|-6arctan(e^x))+C

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integral of (e^x)/((e^x-6)(e^{2x)+1)}

\int \frac{e ^{x}}{( e ^{x} - 6 ) ( e ^{2 x} + 1 )} d x

\frac{1}{37} ln ∣ e^{x} - 6 ∣ + \frac{1}{37} (- \frac{1}{2} ln e^{2 x} + 1 - 6 arctan (e^{x})) + C

Solution steps

\int \frac{e ^{x}}{( e ^{x} - 6 ) ( e ^{2 x} + 1 )} d x

Apply u-substitution

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

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

Take the partial fraction of

\frac{1}{( u - 6 ) ( u ^{2} + 1 )} : \frac{1}{37 ( u - 6 )} + \frac{- u - 6}{37 ( u ^{2} + 1 )}

= \int \frac{1}{37 ( u - 6 )} + \frac{- u - 6}{37 ( u ^{2} + 1 )} d u

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{1}{37 ( u - 6 )} d u + \int \frac{- u - 6}{37 ( u ^{2} + 1 )} d u

\int \frac{1}{37 ( u - 6 )} d u = \frac{1}{37} ln ∣ u - 6 ∣

\int \frac{- u - 6}{37 ( u ^{2} + 1 )} d u = \frac{1}{37} (- \frac{1}{2} ln u^{2} + 1 - 6 arctan (u))

= \frac{1}{37} ln ∣ u - 6 ∣ + \frac{1}{37} (- \frac{1}{2} ln u^{2} + 1 - 6 arctan (u))

Substitute back

u = e^{x}

= \frac{1}{37} ln ∣ e^{x} - 6 ∣ + \frac{1}{37} (- \frac{1}{2} ln (e^{x})^{2} + 1 - 6 arctan (e^{x}))

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= \frac{1}{37} ln ∣ e^{x} - 6 ∣ + \frac{1}{37} (- \frac{1}{2} ln e^{2 x} + 1 - 6 arctan (e^{x}))

Add a constant to the solution

= \frac{1}{37} ln ∣ e^{x} - 6 ∣ + \frac{1}{37} (- \frac{1}{2} ln e^{2 x} + 1 - 6 arctan (e^{x})) + C

f (x) = \frac{1 - ln ( x )}{x ^{2}}

f (x) = \frac{x}{x + 17}

\int 3 x - 2 x d x

\frac{\partial}{\partial y} (y e^{x y})

\frac{d}{d x} (f (x) g (x) h (x))

What is the integral of (e^x)/((e^x-6)(e^{2x)+1)} ?
The integral of (e^x)/((e^x-6)(e^{2x)+1)} is 1/37 ln|e^x-6|+1/37 (-1/2 ln|e^{2x}+1|-6arctan(e^x))+C