# Integrate Using Substitution

• Jun 18th 2007, 06:08 AM
Killswitch
Integrate Using Substitution
I got this question in my maths exam today, I think I got it right - but I'm unsure.

$\displaystyle \int_0^2 \frac{2^x}{(2^x+1)^2} dx$

Use the substitution $\displaystyle u = 2^x$

I got something like 1/ln64.
• Jun 18th 2007, 07:17 AM
ThePerfectHacker
Quote:

Originally Posted by Killswitch
I got this question in my maths exam today, I think I got it right - but I'm unsure.

$\displaystyle \int_0^2 \frac{2^x}{(2^x+1)^2} dx$

Use the substitution $\displaystyle u = 2^x$

I got something like 1/ln64.

$\displaystyle \int_0^2 \frac{e^{x\ln 2}}{\left( e^{x\ln 2} + 1 \right)^2} \ dx$

Let $\displaystyle t = x\ln 2$

$\displaystyle \frac{1}{\ln 2} \int_0^{2\ln 2} \frac{e^t}{(e^t+1)^2} dt$

Let $\displaystyle \chi = e^t+1$

$\displaystyle \int_2^3 \frac{1}{\chi ^2} d \chi$
• Jun 18th 2007, 07:29 AM
Pterid
Yes, that looks like the right substitution... but I get an answer of $\displaystyle \frac{3}{10\ln 2}$.

We have to evaluate $\displaystyle \int_0^2 \frac{2^x}{(2^x+1)^2} dx$ :

Substitute $\displaystyle u = 2^x$

$\displaystyle \Rightarrow \ln u = \ln 2^x = x \ln 2$

$\displaystyle \Rightarrow \frac{1}{u} \frac{du}{dx} = \ln 2$, i.e. $\displaystyle u dx = \frac{du}{\ln 2}$.

So, $\displaystyle \int_0^2 \frac{2^x}{(2^x+1)^2} dx = \int_0^2 \frac{u dx}{(u+1)^2} = \frac{1}{\ln 2} \int_1^4 \frac{du}{(u + 1)^2}$.

$\displaystyle = \frac{1}{\ln 2} \left[ \frac{-1}{(u+1)}\right]_1^4 = \frac{1}{\ln 2} \cdot \left( - \frac{1}{5} + \frac{1}{2}\right) = \frac{3}{10\ln 2}$.
• Jun 18th 2007, 07:47 AM
Plato
Here is another way.
$\displaystyle u = 2^x + 1\quad \Rightarrow \quad du = 2^x \ln (2)dx$
$\displaystyle \int\limits_0^2 {\frac{{2^x dx}}{{\left( {2^x + 1} \right)^2 }} = \frac{1}{{\ln (2)}}\int\limits_2^5 {u^{ - 2} du} }$
• Jun 22nd 2007, 01:29 AM
curvature
Yet another way without changing the upper and lower limits