Question #68d98

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Question #68d98

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Answer 1 · 2017-09-12T19:03:45

The answer is $2 ln (e^{x} + 1) - x + C$ $2ln(e^{x}+1)-x+C$

Explanation:

Start by letting $u = e^{x} + 1$ $u=e^x+1$ so that $d u = e^{x} d x$ $du=e^{x}dx$ and $e^{x} - 1 = u - 2$ $e^x-1=u-2$ . Then $d x = \frac{1}{e^{x}} d u = \frac{1}{u - 2} d u$ $dx=1/e^(x) du=1/(u-2)du$ and the integral becomes

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

Next, use the Method of Partial Fractions to write $\frac{u - 2}{u (u - 1)}$ $(u-2)/(u(u-1))$ as $\frac{2}{u} - \frac{1}{u - 1}$ $2/u-1/(u-1)$ , which can be easily integrated to get $2 ln | u | - ln | u - 1 | + C$ $2ln|u|-ln|u-1|+C$ .

Now substitute $u = e^{x} + 1$ $u=e^x+1$ , which is never negative, and use the fact that $ln (e^{x}) = x$ $ln(e^x)=x$ to get

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

This can be checked by differentiation:

$\frac{d}{d x} (2 ln (e^{x} + 1) - x + C) = \frac{2}{e^{x} + 1} \cdot e^{x} - 1 = \frac{2 e^{x} - (e^{x} + 1)}{e^{x} + 1} = \frac{e^{x} - 1}{e^{x} + 1}$ $d/dx(2ln(e^x+1)-x+C)=2/(e^x+1)*e^x-1=(2e^x-(e^x+1))/(e^x+1)=(e^x-1)/(e^x+1)$

Answer 2 · 2017-09-12T19:14:19

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

Explanation:

Given: $\int \frac{e^{x} - 1}{e^{x} + 1} d x$ $int(e^x-1)/(e^x+1)dx$

Insert zero into the numerator:

$\int \frac{e^{x} + 0 - 1}{e^{x} + 1} d x$ $int(e^x+ 0-1)/(e^x+1)dx$

In place of the 0 we write $e^{x} - e^{x}$ $e^x-e^x$ :

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

Now we do some clever grouping:

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

We combine the first group:

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

Separate into two fractions:

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

Please notice that the second fraction becomes 1:

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

Separate into two integrals:

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

For the first integral we let $u = e^{x} + 1$ $u = e^x+1$ , then $d u = e^{x} d x$ $du = e^xdx$ :

$2 \int \frac{1}{u} d u - \int d x$ $2int1/udu -intdx$

We know these integrals very well:

$2 ln | u | - x + C$ $2ln|u| -x+ C$

Reverse the substitution:

$2 ln (e^{x} + 1) - x + C$ $2ln(e^x+1) -x+ C$

Answer 3 · 2017-09-12T23:17:43

$x + 2 ln (1 + e^{- x}) + C$ $x+2ln(1+e^-x)+C$

Explanation:

We can also do this integral as follows:

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

Splitting up the fraction as $\frac{e^{x} + 1}{e^{x} + 1} - \frac{2}{e^{x} + 1}$ $(e^x+1)/(e^x+1)-2/(e^x+1)$ , this becomes:

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

We can integrate the first term easily:

$= x - 2 \int \frac{1}{e^{x} + 1} d x$ $=x-2int1/(e^x+1)dx$

Now, multiply the integrand by $\frac{e^{- x}}{e^{- x}}$ $e^-x/e^-x$ . This seems ridiculous, but you'll see why it works in a second:

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

Let $u = 1 + e^{- x}$ $u=1+e^-x$ , implying that $d u = - e^{- x} d x$ $du=-e^-xdx$ . Then:

$= x + 2 \int \frac{1}{u} d u$ $=x+2int1/udu$

$= x + 2 ln | u | + C$ $=x+2lnabsu+C$

$= x + 2 ln ∣ ∣ 1 + e^{- x} ∣ ∣ + C$ $=x+2lnabs(1+e^-x)+C$

As $e^{- x} > 0$ $e^-x>0$ for all Real values of $x$ $x$ , the absolute value bars aren't needed.

$x + 2 ln (1 + e^{- x}) + C$ $x+2ln(1+e^-x)+C$

Which can be shown to be equivalent to the other provided answers.

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