$$ \int_0^4 \ln\left(\sqrt{(x-2)^2+4} + (x-2)\right)~{\rm d}x $$
In the above given question, I am unable to integrate the function inside log. I am substituting $(x-2) = t$ and then reducing the question into a simpler form with only $t$ as a variable, but i am getting a function of $t$ inside log which I am unable to open or integrate. Please provide a proper approach to my way or any suggestions are well appreciated.
First, make the substitution you've already mentioned: $$\int_0^4 \ln\left(\sqrt{(x-2)^2+4} + (x - 2)\right)~{\rm d}x$$ $$t=x-2$$ $${\rm d}t={\rm d}x$$ $$\int_{-2}^2 \ln\left(\sqrt{t^2+4} + t\right)~{\rm d}t$$ Now, we solve using integration by parts where $u = \ln\left(\sqrt{t^2+4} + t\right)$ and ${\rm d}v = {\rm d}t$. Calculating further, we get that $v=t$ and $${\rm d}u = \frac{1}{\sqrt{t^2+4} + t} \cdot \frac{\sqrt{t^2+4} + t}{\sqrt{t^2+4}}~{\rm d}t = \frac{{\rm d}t}{\sqrt{t^2+4}}$$ This gives us $$\int_{-2}^2 \ln\left(\sqrt{t^2+4} + t\right)~{\rm d}t = \left.t \ln\left(\sqrt{t^2+4} + t\right)\right\rvert_{-2}^2 - \int_{-2}^2 \frac{t}{\sqrt{t^2+4}}~{\rm d}t$$ Now, we may make the substitutions $w=t^2+4$ and ${\rm d}w = 2t~{\rm d}t$ (which also means $\frac{1}{2}{\rm d}w=t~{\rm d}t$), as well as evaluating the solved part of the integral. $$\int_{-2}^2 \ln\left(\sqrt{t^2+4} + t\right)~{\rm d}t = \left[2\ln(2\sqrt{2} + 2) + 2\ln(2\sqrt{2} - 2)\right] - \int_8^8 \frac{{\rm d}w}{\sqrt{w}}$$ Because the limits of integration on our integral are now equal, we can eliminate it altogether and continue to evaluate. $$\int_{-2}^2 \ln\left(\sqrt{t^2+4} + t\right)~{\rm d}t = 2\left[\ln(2\sqrt{2} + 2) + \ln(2\sqrt{2} - 2)\right]$$ $$= 2\ln[(2\sqrt{2} + 2)(2\sqrt{2} - 2)]$$ $$= 2\ln(8 - 4) = 2\ln4 = \ln\left(4^2\right) = \ln 16 \approx 2.773$$