The value of the integral ∫limits_0^2 (log ≤ft(x^2+2))/((x+2)^2), d x…

$I=\int\limits_0^2 \frac{\log \left(x^2+2\right)}{(x+2)^2} d x$. 

Using integration by parts

Let $f(x) = \log(x^2+2)$

$g(x)=\frac{1}{(x+2)^2}$

Apply integration by part.

$I=\log(x^2+2)\int\frac{1}{(x+2)^2}dx+\int\frac{x}{(x^2+2)}×\frac{1}{(x+2)}dx$

Let $I_1=\int\frac{x}{(x^2+2)(x+2)}$

$\frac{x}{(x^2+2)(x+2)}=\frac{A}{x^2+2}+\frac{B}{x+2}=\frac{Ax+2A+Bx^2+2B}{(x^2+2)(x+2)}$

On comparing we get

$A=\frac{1}{3},B=\frac{-1}{3}$

$⇒I_1=\int\frac{x}{(x^2+2)(x+2)}=\int\frac{1}{3(x^2+2)}dx-\int\frac{1}{3(x+2)}dx$

$=\frac{\sqrt{2}}{3}\tan^{-1}(\frac{x}{\sqrt{2}})-\frac{1}{3}\log|x+2|$

$I=\left[\frac{-\log(x^2+2)}{x+2}+\frac{\sqrt{2}}{3}\tan^{-1}(\frac{x}{\sqrt{2}})-\frac{\log|x+2|}{3}\right]_0^2$

$=\frac{-\log 6}{4}+\frac{\sqrt{2}}{3}\tan^{-1}\sqrt{2}-\frac{\log 4}{3}+\frac{\log 2}{2}+\frac{\log 2}{3}$

$=-\frac{\log 2}{4}-\frac{\log 3}{4}+\frac{\sqrt{2}}{3}\tan^{-1}\sqrt{2}$

$-\frac{2\log 2}{3}+\frac{\log 2}{3}+\frac{\log 2}{2}$

$=\frac{\sqrt{2}}{3}\tan^{-1}\sqrt{2}-\frac{\log 3}{4}+\frac{5}{12}\log2$