\(\frac{1}{1+tanx}=\frac{cosx}{cosx+sinx}=\frac{1}{2}\left(\frac{cosx+sinx}{cosx+sinx}+\frac{-sinx+cosx}{cosx+sinx}\right)\)

\(\Rightarrow\int\limits^{\frac{\pi}{4}}_0\frac{1}{1+tanx}dx=\frac{1}{2}\int\limits^{\frac{\pi}{4}}_0\left(1+\frac{-sinx+cosx}{cosx+sinx}\right)dx=\int\limits^{\frac{\pi}{4}}_0\frac{1}{2}dx+\frac{1}{2}\int\limits^{\frac{\pi}{4}}_0\frac{d\left(cosx+sinx\right)}{cosx+sinx}\)

\(=\frac{1}{2}x|^{\frac{\pi}{4}}_0+\frac{1}{2}ln\left(sinx+cosx\right)|^{\frac{\pi}{4}}_0=\frac{\pi}{8}+\frac{1}{2}ln\sqrt{2}=\frac{\pi}{8}+\frac{1}{4}ln2\)

\(\Rightarrow a=\frac{1}{8}\) ; \(b=\frac{1}{4}\)

Do đó bán kính đường tròn \(\left(S\right)\cap\left(S'\right)\) bằng \(\dfrac{10\sqrt{41}}{41}a\)

a) \(\int\frac{1}{x^2-3x+2}dx=\frac{1}{2-1}\int\frac{1}{\left(x-1\right)\left(x-2\right)}dx\)

=\(\int\frac{1}{x-2}dx-\int\frac{1}{x-1}dx=ln\left|x-2\right|-ln\left|x-1\right|=ln\left|\frac{x-2}{x-1}+C\right|\)

b) \(\int\frac{1}{4x^2-3x-1}dx=\frac{1}{4}.\frac{1}{\left(1-\frac{1}{4}\right)}\int\frac{1}{\left(x+\frac{1}{4}\right)\left(x-1\right)}dx\)

=\(\frac{1}{3}.\left[\int\frac{1}{x-1}dx-\int\frac{1}{x+\frac{1}{4}}dx\right]\)

=\(\frac{1}{3}\left[ln\left|x-1\right|-ln\left|x+\frac{1}{4}\right|\right]=\frac{1}{3}ln\left|\frac{x-1}{x+\frac{1}{4}}\right|+C\)

=\(\frac{1}{3}ln\left|\frac{4\left(x-1\right)}{4x+1}+C\right|\)

Đặt \(t=\sqrt{1-x}\)

\(\Rightarrow\begin{cases}x=1-t^2\\dx=-2tdt\end{cases}\) \(\Leftrightarrow\frac{x^2dx}{\sqrt{1-x}}\)\(=\frac{\left(1-t^2\right)\left(-2tdt\right)}{t}=-2\left(1-2t^2+3t^4-t^6\right)dt\)

Vậy : \(\int\frac{x^3dx}{\sqrt{1-x}}=\int\left(-2+4t^2-6t^4+2t^6\right)dt=-2t+\frac{4}{3}t^3-\frac{6}{5}t^5+\frac{2}{7}t^7+C\)

                    = \(-2\sqrt{1-x}+\frac{4}{3}\left(1-x\right)\sqrt{1-x}-\frac{6}{5}\left(1-x\right)^2\sqrt{1-x}+\frac{2}{7}\left(1-x\right)^3\sqrt{1-x}+C\)

tính E(300)=300/log2(300), E(90000)=90000/log2(90000)

Vì độ hiệu quả tỉ lệ thuận với thời gian thực hiện

nên ta có tỉ số 0,02/E(300)=x/E(90000) (x là giá trị cần tìm).

Từ đó tính được x=3

Ta có :

\(\frac{3x+2}{x^2+2x-3}=\frac{E\left(2x+2\right)+D}{x^2+2x-3}=\frac{2E+D+2E}{x^2+2x-3}\)

Đồng nhất hệ số hai tử sốta có hệ phương trình 

\(\begin{cases}2E=3\\D+2E=2\end{cases}\) \(\Rightarrow\begin{cases}E=\frac{3}{2}\\D=-1\end{cases}\)

\(\Rightarrow\) \(\frac{3x+2}{x^2+2x-3}=\frac{\frac{3}{2}\left(2x+2\right)}{x^2+2x-3}-\frac{1}{x^2+2x-3}\)

Vậy :

\(\int\frac{3x+2}{x^2+2x-3}dx=\frac{3}{2}\int\frac{d\left(x^2+2x-3\right)}{x^2+2x-3}+\int\frac{1}{x^2+2x-3}dx\)\(=\frac{3}{2}\ln\left|x^2+2x-3\right|+J\left(1\right)\)

Tính :

\(J=\int\frac{1}{x^2+2x-3}dx=\frac{1}{4}\left(\int\frac{1}{x-1}dx-\int\frac{1}{x+3}dx\right)=\frac{1}{4}\ln\left|x-1\right|-\ln\left|x+3\right|=\frac{1}{4}\ln\left|\frac{x-1}{x+3}+C\right|\)

Do đó :  \(\int\frac{3x+2}{x^2+2x-3}dx=\frac{3}{2}\ln\left|x^2+2x-3\right|+\frac{1}{4}\ln\left|\frac{x-1}{x+3}\right|+C\)

b) Ta có :

\(\frac{2x-3}{x^2+4x+4}=\frac{E\left(2x+4\right)+D}{x^2+4x+4}=\frac{2Ex+D+4E}{x^2+4x+4}\)

Đồng nhất hệ số  hai tử số :

Ta có hệ : \(\Leftrightarrow\)\(\begin{cases}2E=2\\D+4E=-3\end{cases}\)\(\Leftrightarrow\)\(\begin{cases}E=1\\D=-7\end{cases}\)

Suy ra :

\(\frac{2x-3}{x^2+4x+4}=\frac{2x+4}{x^2+4x+4}-\frac{7}{x^2+4x+4}\)

Vậy : \(\int\frac{2x-3}{x^2+4x+4}dx=\int\frac{2x+4}{x^2+4x+4}dx-7\int\frac{1}{\left(x+2\right)^2}dx=\ln\left|x^2+4x+4\right|+\frac{7}{x+2}+C\)

a) Đặt \(x=\sin t;t\in\left(-\frac{\pi}{2};\frac{\pi}{2}\right)\) \(\Rightarrow dx=\cos tdt\)

Suy ra : \(\frac{dx}{\sqrt{\left(1-x^2\right)^3}}=\frac{\cos tdt}{\sqrt{\left(1-\sin^2t\right)^3}}=\frac{\cos tdt}{\cos^3t}=\frac{dt}{\cos^2t}=d\left(\tan t\right)\)

Khi đó \(\int\frac{dx}{\sqrt{\left(1-x^2\right)^3}}=\int d\left(\tan t\right)=\tan t+C=\frac{\sin t}{\sqrt{1-\sin^2t}}=\frac{x}{\sqrt{1-x^2}}+C\)

b) Vì \(x^2+2x+3=\left(x+1\right)^2+\left(\sqrt{2}\right)^2\)

nên ta đặt : \(x+1=\sqrt{2}\tan t;t\in\left(-\frac{\pi}{2};\frac{\pi}{2}\right)\Rightarrow dx=\sqrt{2}.\frac{dt}{\cos^2t};\tan t=\frac{x+1}{\sqrt{2}}\)

Suy ra \(\frac{dx}{\sqrt{x^2+2x+3}}=\frac{dx}{\sqrt{\left(x^2+1\right)^2+\left(\sqrt{2}\right)^2}}=\frac{dx}{\sqrt{2\left(\tan^2t+1\right).\cos^2t}}\)

                        \(=\frac{dt}{\sqrt{2}\cos t}=\frac{1}{\sqrt{2}}.\frac{\cos tdt}{1-\sin^2t}=-\frac{1}{2\sqrt{2}}.\left(\frac{\cos tdt}{\sin t-1}-\frac{\cos tdt}{\sin t+1}\right)\)

Khi đó \(\int\frac{dx}{\sqrt{x^2+2x+3}}=-\frac{1}{2\sqrt{2}}\int\left(\frac{\cos tdt}{\sin t-1}-\frac{\cos tdt}{\sin t+1}\right)=-\frac{1}{2\sqrt{2}}\ln\left|\frac{\sin t-1}{\sin t+1}\right|+C\left(1\right)\)

Từ \(\tan t=\frac{x+1}{\sqrt{2}}\Leftrightarrow\tan^2t=\frac{\sin^2t}{1-\sin^2t}=\frac{\left(x+1\right)^2}{2}\Rightarrow\sin^2t=1-\frac{2}{x^2+2x+3}\)

Ta tìm được \(\sin t\) thay vào (1), ta tính được I