What is the integral of (e^x)/((e^x-4)(e^{2x)+1)} ?

The integral of (e^x)/((e^x-4)(e^{2x)+1)} is 1/17 ln|e^x-4|+1/17 (-1/2 ln|e^{2x}+1|-4arctan(e^x))+C

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integral of (e^x)/((e^x-4)(e^{2x)+1)}

\int \frac{e ^{x}}{( e ^{x} - 4 ) ( e ^{2 x} + 1 )} d x

\frac{1}{17} ln ∣ e^{x} - 4 ∣ + \frac{1}{17} (- \frac{1}{2} ln e^{2 x} + 1 - 4 arctan (e^{x})) + C

Solution steps

\int \frac{e ^{x}}{( e ^{x} - 4 ) ( e ^{2 x} + 1 )} d x

Apply u-substitution

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

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

Take the partial fraction of

\frac{1}{( u - 4 ) ( u ^{2} + 1 )} : \frac{1}{17 ( u - 4 )} + \frac{- u - 4}{17 ( u ^{2} + 1 )}

= \int \frac{1}{17 ( u - 4 )} + \frac{- u - 4}{17 ( 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}{17 ( u - 4 )} d u + \int \frac{- u - 4}{17 ( u ^{2} + 1 )} d u

\int \frac{1}{17 ( u - 4 )} d u = \frac{1}{17} ln ∣ u - 4 ∣

\int \frac{- u - 4}{17 ( u ^{2} + 1 )} d u = \frac{1}{17} (- \frac{1}{2} ln u^{2} + 1 - 4 arctan (u))

= \frac{1}{17} ln ∣ u - 4 ∣ + \frac{1}{17} (- \frac{1}{2} ln u^{2} + 1 - 4 arctan (u))

Substitute back

u = e^{x}

= \frac{1}{17} ln ∣ e^{x} - 4 ∣ + \frac{1}{17} (- \frac{1}{2} ln (e^{x})^{2} + 1 - 4 arctan (e^{x}))

Apply exponent rule:

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

= \frac{1}{17} ln ∣ e^{x} - 4 ∣ + \frac{1}{17} (- \frac{1}{2} ln e^{2 x} + 1 - 4 arctan (e^{x}))

Add a constant to the solution

= \frac{1}{17} ln ∣ e^{x} - 4 ∣ + \frac{1}{17} (- \frac{1}{2} ln e^{2 x} + 1 - 4 arctan (e^{x})) + C

\frac{d}{d x} (sin (3 x) cos (2 x))

\int \frac{1}{x ^{3} - 1} d x

\frac{d y}{d x}, x^{2} - y^{3} = 6

\int cos (x) + \frac{1}{3} sin (x) - \frac{1}{x ^{2}} d x

What is the integral of (e^x)/((e^x-4)(e^{2x)+1)} ?
The integral of (e^x)/((e^x-4)(e^{2x)+1)} is 1/17 ln|e^x-4|+1/17 (-1/2 ln|e^{2x}+1|-4arctan(e^x))+C