Evaluation of Integrals

**Belowzero78** · December 4th 2009, 07:24 AM

1. Integral of (e^x+1)dx/(e^x-1)

I tried the method of substitution and i let u be e^x-1. then du=e^xdx so solving for dx=du/e^x. I dont think this is correct. It doesnt make anything easier. Can someone lead me in the right direction?

2. Integral of t dt/(sqrt(4-t^4))

I let u = 4-t^4 so du=-4t^3dt, then dt=-du/-4t^3. What do i do now?

**Jhevon** · December 4th 2009, 07:43 AM

Originally Posted by Belowzero78

1. Integral of (e^x+1)dx/(e^x-1)

I tried the method of substitution and i let u be e^x-1. then du=e^xdx so solving for dx=du/e^x. I dont think this is correct. It doesnt make anything easier. Can someone lead me in the right direction?

One of many ways to do this...

$\int \frac {e^x + 1}{e^x - 1}~dx = \int \frac {e^x - 1 + 2}{e^x - 1}~dx$

$= \int \left( 1 + \frac 2{e^x - 1} \right)~dx$

$= x + \int \frac 2{e^x - 1}~dx$

Now consider $\int \frac 2{e^x - 1}~dx$

Let $u = e^x - 1 \implies \boxed{e^x = u + 1}$

$\Rightarrow du = e^x ~dx$

$\Rightarrow \frac {du}{e^x} = dx$

$\Rightarrow \frac {du}{u + 1} = dx$

So the integral becomes:

$2 \int \frac 1{u(u + 1)}~du = 2 \int \frac {u + 1 - u}{u(u + 1)}~du$

$= 2 \int \left( \frac 1u - \frac 1{u + 1} \right)~du$

and I leave you to finish off.

2. Integral of t dt/(sqrt(4-t^4))

I let u = 4-t^4 so du=-4t^3dt, then dt=-du/-4t^3. What do i do now?

Let $u = t^2$

$\Rightarrow du = 2t~dt \implies \frac 12~du = t~dt$

So your integral becomes:

$\frac 12 \int \frac 1{\sqrt{4 - u^2}}~du$

which is of the form $\int \frac 1{\sqrt{a^2 - x^2}}~dx = \sin^{-1} \frac xa + C$

**Soroban** · December 4th 2009, 08:34 AM

Hello, Belowzero78!

$1)\;\;\int\frac{e^x+1}{e^x-1}\,dx$

Here's a back-door approach . . .

Divide top and bottom by $e^{\frac{x}{2}}$ :

. . $\int \frac{\frac{1}{e^{\frac{x}{2}}}(e^x + 1)} {\frac{1}{e^{\frac{x}{2}}}(e^x-1)} \,dx \;=\;\int\frac{(e^{\frac{x}{2}} + e^{-\frac{x}{2}})\,dx}{e^{\frac{x}{2}}-e^{-\frac{x}{2}}}$

Let: . $u \:=\:e^{\frac{x}{2}} - e^{-\frac{x}{2}} \quad\Rightarrow\quad du \:=\:\left(\tfrac{1}{2}e^{\frac{x}{2}} + \tfrac{1}{2}e^{-\frac{x}{2}}\right)\,dx$ . $\Rightarrow \quad\left(e^{\frac{x}{2}} + e^{-\frac{x}{2}}\right)\,dx \:=\:2\,du$

Substitute: . $\int\frac{2\,du}{u} \;\;=\;\;2\ln|u| + C \quad\Rightarrow\quad 2\ln\left|e^{\frac{x}{2}} - e^{-\frac{x}{2}}\right| + C$

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