Integral of $e^x \, \ln(e^{2x} - 4)$

Question

Find the integral from $\ln 4$ to $\ln 6$ of $$e^x \ln(e^{2x} - 4)$$

I factored $$\ln(e^{2x} - 4)$$ to get $$\ln((e^{x} - 2)(e^{x} + 2))$$

Then I separated this to get:

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

So then I integrated to get $$(1/2) (\ln(e^{x} - 2))^2 + (1/2)(\ln(e^{x} + 2))^2$$

Then I attempted to evaluate the integral, but I didn't know how to simplify for example $$(\ln2)^2$$ And the available options (this is multiple choice) did not contain any such squared values. Is there a way to simplify this? Or should the question be done differently?

Answer 1

Let $$I = \int_{\ln(4)}^{\ln(6)} e^x \ln(e^{2x}-4)dx$$ Now set $t=e^x$ and we get $$I = \int_4^6 \ln(t^2-4) dt = \int_4^6 \ln(t+2) dt + \int_4^6 \ln(t-2) dt = \int_6^8 \ln(t)dt + \int_2^4 \ln(t)dt$$ I trust you can finish it from here.

Answer 2

If the answers don't look like yours, there's chance that you've done something wrong. The integral of $\ln x$ is not $\frac{1}{2} \ln^2 x$. It's evident when you take the derivative:

$$ \frac{d}{dx} \left( \frac{1}{2} \ln^2 x \right) = \frac{\ln x}{x} $$

For the $\ln$ function, try integration by parts. Also, use a substitution first, don't think you can do everything in your head.

Let $t = e^x$,

$$ \int_{\ln 4}^{\ln 6} e^x \ln (e^x - 2) \, dx = \int_4^6 \ln(t-2) \, dt $$

Let $$u = \ln (t-2)$$ $$dv = dt$$

You'll get $$ du = \frac{dt}{t-2} $$ $$ v = t - 2 $$

So

$$ \begin{align*} \int_4^6 \ln(t-2) \, dt &= \int u\,dv \\ &= uv - \int v\,du \\ &= (t-2) \ln(t-2) \bigg|_4^6 - \int_4^6 dt \\ &= (t-2) \ln(t-2) \bigg|_4^6 - t \bigg|_4^6 \end{align*} $$

Do the same for the other integrals.