# Thread: Integrals = no mercy

1. ## Integrals = no mercy

5) (i) Use the substitution $u^2=2x+1$ to show that, $x>4$

$\int\frac{3}{(x-4)\sqrt{2x+1}}dx=\ln(\frac{\sqrt{2x+1}-3}{\sqrt{2x+1}+3})+K$

where K is a constant.

(ii) Show that (from ln 8 to ln 3) $\int\frac{2}{e^x\sqrt{e^x+1}}=\frac{7}{12}+\ln\fra c{2}{3}$

2. Originally Posted by iPod
(ii) Show that (from ln 8 to ln 3) $\int\frac{2}{e^x\sqrt{e^x+1}}=\frac{7}{12}+\ln\fra c{2}{3}$

For

$\int_{ln(3)}^{ln(8)}\frac{2}{e^x\sqrt{e^x+1}}dx$

consider $u = e^x+1 \Rightarrow \frac{du}{dx}= e^x$

3. Originally Posted by iPod

5) (i) Use the substitution $u^2=2x+1$ to show that, $x>4$

$\int\frac{3}{(x-4)\sqrt{2x+1}}dx=\ln(\frac{\sqrt{2x+1}-3}{\sqrt{2x+1}+3})+K$
Hint is given, I think you can show what you've done.

4. Originally Posted by iPod
5) (i) Use the substitution $u^2=2x+1$ to show that, $x>4$

$\int\frac{3}{(x-4)\sqrt{2x+1}}dx=\ln(\frac{\sqrt{2x+1}-3}{\sqrt{2x+1}+3})+K$

where K is a constant.

(ii) Show that (from ln 8 to ln 3) $\int\frac{2}{e^x\sqrt{e^x+1}}=\frac{7}{12}+\ln\fra c{2}{3}$

For the first, you have to let $u = \sqrt{2x + 1}$.

This means that $\frac{du}{dx} = \frac{1}{\sqrt{2x + 1}} = \frac{1}{u}$.

With some algebraic manipulation, you can see that

$u^2 = 2x + 1$

$x = \frac{1}{2}u^2 - \frac{1}{2}$

Therefore $x - 4 = \frac{1}{2}u^2 - \frac{9}{2}$.

So $\frac{3}{(x - 4)\sqrt{2x + 1}} = \frac{3}{x - 4}\,\frac{1}{\sqrt{2x + 1}}$

$= \frac{3}{\frac{1}{2}u^2 - \frac{9}{2}} \, \frac{du}{dx}$

$= \frac{6}{u^2 - 9}\,\frac{du}{dx}$

$= \frac{6}{(u - 3)(u + 3)}\,\frac{du}{dx}$.

Can you go from here? You need to use partial fractions.