How to integrate #int(e^2x+e^-2x)/(e^2x-e^-2x)dx?

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How to integrate #int(e^2x+e^-2x)/(e^2x-e^-2x)dx?

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Answer 1 · 2018-05-23T15:28:46

We can rewrite as

$I = \int \frac{e^{2 x} + \frac{1}{e^{2 x}}}{e^{2 x} - \frac{1}{e^{2 x}}} d x$ $I = int (e^(2x) + 1/e^(2x))/(e^(2x) - 1/e^(2x)) dx$

$I = \int \frac{\frac{e^{4 x} + 1}{e^{2 x}}}{\frac{e^{4 x} - 1}{e^{2 x}}} d x$ $I = int((e^(4x) + 1)/e^(2x))/((e^(4x) - 1)/e^(2x)) dx$

$I = \int \frac{e^{4 x} + 1}{e^{4 x} - 1} d x$ $I = int(e^(4x) + 1)/(e^(4x) - 1)dx$

Let $u = 4 x$ $u = 4x$ . Then $d u = 4 d x$ $du = 4dx$ and $d x = \frac{d u}{4}$ $dx= (du)/4$

$I = \frac{1}{4} \int \frac{e^{u} + 1}{e^{u} - 1} d u$ $I = 1/4 int (e^u + 1)/(e^u - 1)du$

Now use partial fractions.

$\frac{A}{e^{u} - 1} + \frac{B}{1} = \frac{e^{u} + 1}{e^{u} - 1}$ $A/(e^u - 1) + B/1 = (e^u + 1)/(e^u - 1)$

$A + B e^{u} - B = e^{u} + 1$ $A + Be^u - B = e^u + 1$

We can readily see that $B = 1$ $B = 1$ and therefore $A = 2$ $A = 2$ .

Thus

$I = \frac{1}{4} (\int \frac{2}{e^{u} - 1} + 1 d u)$ $I = 1/4(int 2/(e^u - 1) + 1 du)$

$I = \frac{1}{2} ln | e^{u} - 1 | + \frac{1}{4} u + C$ $I = 1/2ln|e^u - 1| + 1/4u + C$

$I = \frac{1}{2} ln ∣ ∣ e^{4 x} - 1 ∣ ∣ + x + C$ $I = 1/2ln|e^(4x) - 1| + x + C$

Hopefully this helps!

Answer 2 · 2018-05-23T16:31:55

The answer is $= \frac{1}{2} ln (\frac{1}{2} (∣ ∣ e^{2 x} - e^{- (2 x)} ∣ ∣) + C$ $=1/2ln(1/2(|e^(2x)-e^-(2x)|)+C$

Explanation:

The function is

$\frac{(e^{2 x} - e^{- (2 x)})}{(e^{2 x} - e^{- (2 x)})} = coth (2 x) = \frac{cosh (2 x)}{sinh (2 x)}$ $((e^(2x)-e^-(2x)))/((e^(2x)-e^-(2x)))=coth(2x)=cosh(2x)/sinh(2x)$

Therefore, the integral is

$I = \int \frac{(e^{2 x} - e^{- (2 x)}) d x}{e^{2 x} - e^{- (2 x)}}$ $I=int((e^(2x)-e^-(2x))dx)/(e^(2x)-e^-(2x))$

$= \int coth (2 x) d x$ $=intcoth(2x)dx$

$= \int \frac{cosh (2 x) d x}{sinh (2 x)}$ $=int(cosh(2x)dx)/sinh(2x)$

Let $u = sinh (2 x)$ $u=sinh(2x)$ , $\Rightarrow$ $=>$ , $d u = 2 cosh (2 x) d x$ $du=2cosh(2x)dx$

So, the integral is

$I = \frac{1}{2} \int \frac{d u}{u}$ $I=1/2int(du)/(u)$

$= \frac{1}{2} ln (u)$ $=1/2ln(u)$

$= \frac{1}{2} ln (sinh (2 x)) + C$ $=1/2ln(sinh(2x))+C$

$= \frac{1}{2} ln (\frac{1}{2} (∣ ∣ e^{2 x} - e^{- (2 x)} ∣ ∣) + C$ $=1/2ln(1/2(|e^(2x)-e^-(2x)|)+C$

Answer 3 · 2018-05-23T16:43:39

We have,

$\int \frac{e^{2 x} + e^{- 2 x}}{e^{2 x} - e^{- 2 x}} d x$ $int (e^(2x) + e^(-2x))/(e^(2x) - e^(-2x))dx$

Let's Substitute $u = e^{2 x} - e^{- 2 x}$ $u = e^(2x) - e^(-2x)$ .

So, $d u = 2 e^{2 x} - (- 2 e^{- 2 x}) d x \Rightarrow d u = 2 e^{2 x} + 2 e^{- 2 x} d x$ $du = 2e^(2x) - (-2e^(-2x))dx rArr du = 2e^(2x) + 2e^(-2x)dx$

$\Rightarrow d x = \frac{1}{2 e^{2 x} + 2 e^{- 2 x}} d u$ $rArr dx = 1/(2e^(2x) + 2e^(-2x)) du$

So,

$\int \frac{e^{2 x} + e^{- 2 x}}{e^{2 x} - e^{- 2 x}} d x$ $int (e^(2x) + e^(-2x))/(e^(2x) - e^(-2x))dx$

$= int cancel(e^(2x) + e^(-2x))/u xx (1/(2cancel((e^(2x) + e^(-2x))))) du$

$= 1/2 int 1/(u) du$

$= 1/2 ln|u| + C$

$= 1/2 ln|e^(2x) - e^(-2x)| + C$

$= 1/2 ln|e^(-2x)(e^(4x) - 1)| + C$

$= 1/2 ln |e^(-2x)| + 1/2 ln|e^(4x) - 1| + C$

$= 1/2 xx -2x + 1/2 ln|e^(4x) - 1| + C$

$= 1/2 ln|e^(4x) - 1| - x + C$

Hope this helps.

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How to integrate #int(e^2x+e^-2x)/(e^2x-e^-2x)dx?

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