# Math Help - Integrate Using Substitution

1. ## Integrate Using Substitution

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

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

Use the substitution $u = 2^x$

I got something like 1/ln64.

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

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

Use the substitution $u = 2^x$

I got something like 1/ln64.
$\int_0^2 \frac{e^{x\ln 2}}{\left( e^{x\ln 2} + 1 \right)^2} \ dx$

Let $t = x\ln 2$

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

Let $\chi = e^t+1$

$\int_2^3 \frac{1}{\chi ^2} d \chi$

3. Yes, that looks like the right substitution... but I get an answer of $\frac{3}{10\ln 2}$.

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

Substitute $u = 2^x$

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

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

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

$= \frac{1}{\ln 2} \left[ \frac{-1}{(u+1)}\right]_1^4 = \frac{1}{\ln 2} \cdot \left( - \frac{1}{5} + \frac{1}{2}\right) = \frac{3}{10\ln 2}$.

4. Here is another way.
$u = 2^x + 1\quad \Rightarrow \quad du = 2^x \ln (2)dx$
$\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} }
$

5. Yet another way without changing the upper and lower limits