$\int\limits_{2}^{+\infty }\left (\frac{1}{x^2-1}+\frac{2}{(x+1)^2} \right )dx=\int\limits_{2}^{+\infty }\left (\frac{1}{2(x-1)}-\frac{1}{2(x+1)}+\frac{2}{(x+1)^2} \right )dx$
 $=\left[ {\frac{1}{2}\ln|x-1|-\frac{1}{2}\ln|x+1|-\frac{2}{x+1}} \right]_{2}^{+\infty }=\left[ {\frac{1}{2}\ln\dfrac{|x-1|}{|x+1}-\frac{2}{x+1}} \right]_{2}^{+\infty }=\dfrac{2}{3}+\dfrac{\ln 3}{2}$

$\int\limits_{\sqrt{2}}^{+\infty }\dfrac{x dx}{(x^2+1)^3}=\dfrac{ 1}{2}\int\limits_{\sqrt{2}}^{+\infty }\dfrac{ d(x^2+1)}{(x^2+1)^3}=\left[ {-\dfrac{ 1}{4}\dfrac{1}{(x^2+1)^2}} \right]_{\sqrt{2}}^{+\infty }=\dfrac{1}{36}$
Chú ý rằng
$\lim_{x \to +\infty}\dfrac{1}{(x^2+1)^2}=0$

Ta có
$\dfrac{\arctan x}{x^2}=-\dfrac{ x}{x^2+1}+\dfrac{ 1}{x}-\dfrac{\dfrac{ x}{x^2+1}-\arctan x}{x^2}$
$\dfrac{\arctan x}{x^2}=\left (-\dfrac{ 1}{2}\ln(x^2+1) \right )'+\left (\ln x \right )'-\left (\dfrac{\arctan x}{x} \right )'$
$\dfrac{\arctan x}{x^2}=\left (\ln\dfrac{ x}{\sqrt{x^2+1}} \right )'-\left (\dfrac{\arctan x}{x} \right )'$
 Suy ra
$\int\limits_{1}^{+\infty }\dfrac{\arctan x}{x^2}dx=\left[ {\ln\dfrac{ x}{\sqrt{x^2+1}}-\dfrac{\arctan x}{x}} \right]_{1}^{+\infty }=\dfrac{\pi+\ln 4}{4}$
Chú ý rằng
$\lim_{x \to +\infty}\ln\dfrac{ x}{\sqrt{x^2+1}} =\ln\left (  \lim_{x \to +\infty}\dfrac{ x}{\sqrt{x^2+1}} \right )=\ln 1 =0$
$\lim_{x \to +\infty}\dfrac{\arctan x}{x} \underbrace{=}_{L'Hospital}\lim_{x \to +\infty}\dfrac{(\arctan x)'}{(x)'}=\lim_{x \to +\infty}\dfrac{\dfrac{1}{x^2+1}}{1} =0$

$I=\int\limits_{-\infty }^{+\infty }\dfrac{dx}{x^2+4x+9}=\int\limits_{-\infty }^{+\infty }\dfrac{d(x+2)}{(x+2)^2+5}=\left[ {\dfrac{1}{\sqrt 5}\arctan\dfrac{x+2}{\sqrt 5}} \right]_{-\infty }^{+\infty }=\dfrac{\pi}{\sqrt 5}$
 Trong đó đã sử dụng
$\arctan(+\infty)=\dfrac{\pi}{2},\arctan(-\infty)=\dfrac{-\pi}{2}$

Nếu bạn là học sinh lớp $9$ thì có thể làm theo các bước sau đây

+ Kẻ $BH \perp AC, DK \perp AC.$ Chứng minh $\triangle ABH=\triangle CDK (ch.gn)\Rightarrow AH=CK.$

+ Chứng minh $\triangle ABH \sim \triangle ACE (g.g) \Rightarrow \dfrac{AB}{AC}=\dfrac{AH}{AE}=\dfrac{CK}{AE}\Rightarrow AB.AE=AC.CK$

+ Chứng minh $\triangle ADK \sim \triangle ACF (g.g) \Rightarrow \dfrac{AD}{AC}=\dfrac{AK}{AF}\Rightarrow AD.AF=AC.AK$

 Vậy
$AB.AE+AD.AF=AC.(CK+AK)=AC.AC=AC^2$

Nếu bạn là học sinh lớp $10$. Bạn có thể xem đáp án tại

http://toan.hoctainha.vn/Thu-Vien/Bai-Tap/103441/bai-103441