91Ó°ÊÓ

The given integral is, $$\int_{0}^{2} \frac{x}{x^{2}+4 x+8} dx$$ Here, our function to be integrated is, $$f(x) = \frac{x}{x^{2}+4 x+8}$$

Now we need to integrate the function \(f(x)\). To do so, we need to find the antiderivative of \(f(x)\). Observe that we can complete the square in the denominator: $$x^2 + 4x + 8 = (x^2+4x+4) + 4 = (x+2)^2 + 4$$ Letting \(u = x + 2\), \(du = dx\), the integral becomes: $$\int_{0}^{2} \frac{x}{(x+2)^2+4}dx = \int_{-2}^{0}\frac{u-2}{u^2+4}du$$ Keep in mind this substitution is not strictly required, but it does simplify the presentation of the integral. Now we can use substitution method, let's consider, $$v = u^{2}+4$$ $$dv = 2udu$$ Now we have, $$\int_{-2}^{0} \frac{1}{2}\frac{u-2}{v} dv$$ We can notice that the integral is of the form \(\int \frac{u}{v} dv\), now integrate using partial fractions: $$\int_{-2}^{0} \frac{1}{2}\frac{u-2}{v} dv = \int_{-2}^{0} \frac{1}{2}\left(\frac{u}{v}-\frac{2}{v}\right) dv$$ $$ = \frac{1}{2} \int_{-2}^{0} \frac{u}{v} dv -\frac{1}{2}\int_{-2}^{0} \frac{2}{v}dv $$ Now, integrating both parts: $$= \frac{1}{2} [\ln|v|]_{-2}^{0} - \frac{1}{2} [2\arctan(u)]_{-2}^{0}$$ We substitute \(u\) and \(v\) values back in: $$=\frac{1}{2}[\ln(u^2+4)]_{-2}^{0} - \frac{1}{2}[2\arctan(u)]_{-2}^{0}$$

Finally, we will apply the given integration limits 0 and 2 to the antiderivative we found: $$= \frac{1}{2}[\ln(2^2+4) - \ln((-2)^2+4)] - \frac{1}{2}\left[2\arctan(0)- 2\arctan(-2)\right]$$ After simplifying the expression we get, $$=\ln{\sqrt{2}} - (\arctan(0)- \arctan(-2))= \boxed{\ln \sqrt{2} + \arctan(2)}$$ Hence, the integral \(\int_{0}^{2} \frac{x}{x^{2}+4 x+8} dx = \ln \sqrt{2} + \arctan(2)\).