أتدرب وأحل المسائل

التكامل بالكسور الجزئية

أجد كلاً من التكاملات الآتية:

$\int \frac{x-10}{x(x+5)}dx$ (1)

$\begin{array}{rl}& \frac{x-10}{x(x+5)}=\frac{A}{x}+\frac{B}{x+5}\\ & \Rightarrow x-10=A(x+5)+Bx\\ & x=0\Rightarrow A=-2\\ & \begin{array}{r}x=-5\Rightarrow B=3\end{array}\\ & \int \frac{x-10}{x(x+5)}dx=\int (\frac{-2}{x}+\frac{3}{x+5})dx\\ & =-2\ln \left|x\right|+3\ln |x+5|+C\end{array}$

$\int \frac{2}{1-x^2}dx$ (2)

$\begin{array}{rl}& \frac{2}{1-x^2}=\frac{2}{(1-x)(1+x)}=\frac{A}{1-x}+\frac{B}{1+x}\\ & \Rightarrow 2=A(1+x)+B(1-x)\\ & x=1\Rightarrow A=1\\ & \begin{array}{rl}x=-1\Rightarrow B& =1\\ \int \frac{2}{1-x^2}dx& =\int (\frac{1}{1-x}+\frac{1}{1+x})dx\\ & =-\ln |1-x|+\ln |1+x|+C\\ & =\ln \left|\frac{1+x}{1-x}\right|+C\end{array}\end{array}$

$\int \frac{4}{(x-2)(x-4)}dx$ (3)

$\begin{array}{rl}& \frac{4}{(x-2)(x-4)}=\frac{A}{x-2}+\frac{B}{x-4}\\ & \Rightarrow 4=A(x-4)+B(x-2)\\ & x=2\Rightarrow A=-2\\ & x=4\Rightarrow B=2\\ & \int \frac{4}{(x-2)(x-4)}dx=\int (\frac{-2}{x-2}+\frac{2}{x-4})dx\\ & =-2\ln |x-2|+2\ln |x-4|+C\\ & =2\ln \left|\frac{x-4}{x-2}\right|+C\end{array}$

$\int \frac{3x+4}{x^2+x}dx$ (4)

$\begin{array}{rl}& \frac{3x+4}{x^2+x}=\frac{3x+4}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1}\\ & \Rightarrow 3x+4=A(x+1)+Bx\\ & x=0\Rightarrow A=4\\ & x=-1\Rightarrow B=-1\\ & \begin{array}{rl}\frac{3x+4}{x^2+x}dx& =\int (\frac{4}{x}+\frac{-1}{x+1})dx\\ & =4\ln \left|x\right|-\ln |x+1|+C\end{array}\end{array}$

$\int \frac{x^2}{x^2-4}dx$ (5)

$\begin{array}{rl}& \int \frac{x^2}{x^2-4}dx=\int (1+\frac{4}{x^2-4})dx\\ & \frac{4}{x^2-4}=\frac{4}{(x-2)(x+2)}=\frac{A}{x-2}+\frac{B}{x+2}\\ & \Rightarrow 4=A(x+2)+B(x-2)\\ & \begin{array}{rl}x=2\Rightarrow A& =1\\ x=-2\Rightarrow B& =-1\\ \int \frac{x^2}{x^2-4}dx& =\int (1+\frac{1}{x-2}+\frac{-1}{x+2})dx\\ & =x+ln|x-2|-ln|x+2|+C\\ & =x+ln\left|\frac{x-2}{x+2}\right|+C\end{array}\end{array}$

$\int \frac{3x-6}{x^2+x-2}dx$ (6)

$\begin{array}{rl}& \frac{3x-6}{x^2+x-2}=\frac{3x-6}{(x+2)(x-1)}=\frac{A}{x+2}+\frac{B}{x-1}\\ & \Rightarrow 3x-6=A(x-1)+B(x+2)\\ & x=-2\Rightarrow A=4\\ & x=1\Rightarrow B=-1\\ & \begin{array}{rl}& x\frac{3x-6}{x^2+x-2}dx=\int (\frac{4}{x+2}+\frac{-1}{x-1})dx\\ & =4\ln |x+2|-\ln |x-1|+C\end{array}\end{array}$

$\int \frac{4x+10}{4x^2-4x-3}dx$ (7)

$\begin{array}{rl}& \frac{4x+10}{4x^2-4x-3}=\frac{4x+10}{(2x-3)(2x+1)}=\frac{A}{2x-3}+\frac{B}{2x+1}\\ & \Rightarrow 4x+10=A(2x+1)+B(2x-3)\\ & x=\frac{3}{2}\Rightarrow A=4\\ & x=-\frac{1}{2}\Rightarrow B=-2\\ & \int \frac{4x+10}{4x^2-4x-3}dx=\int (\frac{4}{2x-3}+\frac{-2}{2x+1})dx\\ & =2\ln |2x-3|-\ln |2x+1|+C\end{array}$

$\int \frac{2x^2+9x-11}{x^3+2x^2-5x-6}dx$ (8)

$\begin{array}{rl}& \frac{2x^2+9x-11}{x^3+2x^2-5x-6}=\frac{2x^2+9x-11}{(x-2)(x+1)(x+3)}=\frac{A}{x-2}+\frac{B}{x+1}+\frac{C}{x+3}\\ & \Rightarrow 2x^2+9x-11=A(x+1)(x+3)+B(x-2)(x+3)+C(x-2)(x+1)\\ & x=2\Rightarrow A=1\\ & x=-1\Rightarrow B=3\\ & x=-3\Rightarrow C=-2\\ & \int \frac{2x^2+9x-11}{x^3+2x^2-5x-6}dx=\int (\frac{1}{x-2}+\frac{3}{x+1}+\frac{-2}{x+3})dx\\ & =\ln |x-2|+3\ln |x+1|-2\ln |x+3|+C\end{array}$

$\int \frac{4x}{x^2-2x-3}dx$ (9)

$\begin{array}{rl}& \frac{4x}{x^2-2x-3}=\frac{4x}{(x-3)(x+1)}=\frac{A}{x-3}+\frac{B}{x+1}\\ & \Rightarrow 4x=A(x+1)+B(x-3)\\ & x=3\Rightarrow A=3\\ & x=-1\Rightarrow B=1\\ & \int \frac{4x}{x^2-2x-3}dx=\int (\frac{3}{x-3}+\frac{1}{x+1})dx\\ & =3\ln |x-3|+\ln |x+1|+C\end{array}$

$\int \frac{8x^2-19x+1}{(2x+1)(x-2{)}^2}dx$ (10)

$\begin{array}{rl}& \frac{8x^2-19x+1}{(2x+1)(x-2{)}^2}=\frac{A}{2x+1}+\frac{B}{x-2}+\frac{C}{(x-2{)}^2}\\ & \Rightarrow 8x^2-19x+1=A(x-2{)}^2+B(2x+1\left)\right(x-2)+C(2x+1)\\ & x=-\frac{1}{2}\Rightarrow A=2\\ & x=2\Rightarrow C=-1\\ & x=0\Rightarrow 1=4A-2B+C\Rightarrow B=3\\ & \int \frac{8x^2-19x+1}{(2x+1)(x-2{)}^2}dx=\int (\frac{2}{2x+1}+\frac{3}{x-2}+\frac{-1}{(x-2{)}^2})dx\\ & =\ln |2x+1|+3\ln |x-2|+\frac{1}{x-2}+C\end{array}$

$\int \frac{9x^2-3x+2}{9x^2-4}dx$ (11)

$\begin{array}{rl}& \int \frac{9x^2-3x+2}{9x^2-4}dx=\int (1+\frac{6-3x}{9x^2-4})dx\\ & \frac{6-3x}{9x^2-4}=\frac{6-3x}{(3x-2)(3x+2)}=\frac{A}{3x-2}+\frac{B}{3x+2}\\ & \Rightarrow 6-3x=A(3x+2)+B(3x-2)\\ & x=\frac{2}{3}\Rightarrow A=1\\ & x=-\frac{2}{3}\Rightarrow B=-2\\ & \int \frac{9x^2-3x+2}{9x^2-4}dx=\int (1+\frac{1}{3x-2}+\frac{-2}{3x+2})dx\\ & =x+\frac{1}{3}\ln |3x-2|-\frac{2}{3}\ln |3x+2|+C\end{array}$

$\int \frac{x^3+2x^2+2}{x^2+x}dx$ (12)

$\begin{array}{rl}& \int \frac{x^3+2x^2+2}{x^2+x}dx=\int (x+1+\frac{2-x}{x^2+x})dx\\ & \frac{2-x}{x^2+x}=\frac{2-x}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1}\\ & \Rightarrow 2-x=A(x+1)+Bx\\ & x=0\Rightarrow A=2\\ & x=-1\Rightarrow B=-3\\ & \int \frac{x^3+2x^2+2}{x^2+x}dx=\int (x+1+\frac{2}{x}+\frac{-3}{x+1})dx\\ & =\frac{1}{2}{x}^2+x+2\ln \left|x\right|-3\ln |x+1|+C\end{array}$

$\int \frac{x^2+x+2}{3-2x-x^2}dx$ (13)

$\begin{array}{rl}& \int \frac{x^2+x+2}{3-2x-x^2}dx=\int (-1+\frac{5-x}{-x^2-2x+3})dx\\ & \frac{5-x}{-x^2-2x+3}=\frac{x-5}{(x+3)(x-1)}=\frac{A}{x+3}+\frac{B}{x-1}\\ & \Rightarrow x-5=A(x-1)+B(x+3)\\ & x=-3\Rightarrow A=2\\ & \begin{array}{r}x=1\Rightarrow B=-1\end{array}\\ & \int \frac{x^2+x+2}{3-2x-x^2}dx=\int (-1+\frac{2}{x+3}+\frac{-1}{x-1})dx\\ & =-x+2\ln |x+3|-\ln |x-1|+C\end{array}$

$\int \frac{2x-4}{(x^2+4)(x+2)}dx$ (14)

$\begin{array}{rl}& \frac{2x-4}{(x^2+4)(x+2)}=\frac{A}{x+2}+\frac{Bx+C}{x^2+4}\\ & \Rightarrow 2x-4=A(x^2+4)+(Bx+C)(x+2)\\ & x=-2\Rightarrow A=-1\\ & x=0\Rightarrow -4=4A+2C\Rightarrow C=0\\ & x=1\Rightarrow -2=5A+3B+3C\Rightarrow B=1\\ & \int \frac{2x-4}{(x^2+4)(x+2)}dx=\int (\frac{-1}{x+2}+\frac{x}{x^2+4})dx\\ & =-\ln |x+2|+\frac{1}{2}\ln |x^2+4|+C\end{array}$

$\int \frac{x^3-4x^2-2}{x^3+x^2}dx$ (15)

$\begin{array}{rl}& \int \frac{x^3-4x^2-2}{x^3+x^2}dx=\int (1+\frac{-5x^2-2}{x^3+x^2})dx\\ & \frac{-5x^2-2}{x^3+x^2}=\frac{-5x^2-2}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}\\ & \Rightarrow -5x^2-2=Ax(x+1)+B(x+1)+C{x}^2\\ & x=0\Rightarrow B=-2\\ & x=-1\Rightarrow C=-7\\ & x=1\Rightarrow -7=2A+2B+C\Rightarrow A=2\\ & \int \frac{x^3-4x^2-2}{x^3+x^2}dx=\int (1+\frac{2}{x}+\frac{-2}{x^2}+\frac{-7}{x+1})dx\\ & =x+2\ln \left|x\right|+\frac{2}{x}-7\ln |x+1|+C\end{array}$

$\int \frac{3-x}{2-5x-12x^2}dx$ (16)

$\begin{array}{rl}& \frac{3-x}{2-5x-12x^2}=\frac{x-3}{12x^2+5x-2}=\frac{x-3}{(4x-1)(3x+2)}=\frac{A}{4x-1}+\frac{B}{3x+2}\\ & \Rightarrow x-3=A(3x+2)+B(4x-1)\\ & x=\frac{1}{4}\Rightarrow A=-1\\ & x=-\frac{2}{3}\Rightarrow B=1\\ & \int \frac{3-x}{2-5x-12x^2}dx=\int (\frac{-1}{4x-1}+\frac{1}{3x+2})dx\\ & =-\frac{1}{4}\ln |4x-1|+\frac{1}{3}\ln |3x+2|+C\end{array}$

$\int \frac{3x^3-x^2+12x-6}{x^4+6x^2}dx$ (17)

$\begin{array}{rl}& \frac{3x^3-x^2+12x-6}{x^4+6x^2}=\frac{3x^3-x^2+12x-6}{x^2(x^2+6)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+6}\\ & \Rightarrow 3x^3-x^2+12x-6=Ax(x^2+6)+B(x^2+6)+(Cx+D)\left(x^2\right)\\ & x=0\Rightarrow B=-1\\ & x=1\Rightarrow 8=7A+7B+C+D\dots \dots \dots \dots \dots \left(1\right)\\ & x=-1\Rightarrow -22=-7A+7B-C+D\dots \dots \left(2\right)\\ & x=2\Rightarrow 38=20A+10B+8C+4D\dots \dots \text{(3)}\end{array}$

 بجمع (1)، (2) ينتج أن: $14B+2D=-14$، وبتعويض $B=-1$، نجد أن $D=0$

وبطرح (2) من (1) ينتج أن $14A+2C=30$ أي أن $C=15-7A$

بالتعويض في (3) ينتج أن: 

$$\begin{gathered} \begin{array}{rl}20A-10+8(15-7A)& =38\\ -36A& =-72\Rightarrow A=2\\ C& =15-7\left(2\right)=1\end{array} \\ \begin{array}{r}\int \frac{3x^3-x^2+12x-6}{x^4+6x^2}dx=\int (\frac{2}{x}+\frac{-1}{x^2}+\frac{x}{x^2+6})dx\\ =2ln\left|x\right|+\frac{1}{x}+\frac{1}{2}ln|x^2+6|+C\end{array} \end{gathered}$$

$\int \frac{5x-2}{(x-2{)}^2}dx$ (18)

$\begin{array}{rl}& \frac{5x-2}{(x-2{)}^2}=\frac{A}{x-2}+\frac{B}{(x-2{)}^2}\\ & \Rightarrow 5x-2=A(x-2)+B\\ & x=2\Rightarrow B=8\\ & x=0\Rightarrow -2=-2A+B\Rightarrow A=5\\ & \int \frac{5x-2}{(x-2{)}^2}dx=\int (\frac{5}{x-2}+\frac{8}{(x-2{)}^2})dx\\ & =5\ln |x-2|-\frac{8}{x-2}+C\end{array}$

ملاحظة: يمكن حل هذا التكامل بالتعويض $u=x-2$

كما يمكن حله بالأجزاء حيث: $u=5x-2,dv=(x-2{)}^{-2}$

أجد قيمة كل من التكاملات الآتية:

${\int }_2^4\frac{6+3x-x^2}{x^3+2x^2}dx$ (19)

$\begin{array}{rl}& \frac{6+3x-x^2}{x^3+2x^2}=\frac{6+3x-x^2}{x^2(x+2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+2}\\ & \Rightarrow 6+3x-x^2=Ax(x+2)+B(x+2)+C\left(x^2\right)\\ & x=0\Rightarrow B=3\\ & x=-2\Rightarrow C=-1\\ & x=1\Rightarrow 8=3A+3B+C\Rightarrow A=0\\ & \begin{array}{rl}{\int }_2^4\frac{6+3x-x^2}{x^3+2x^2}dx& ={\int }_2^4(\frac{3}{x^2}+\frac{-1}{x+2})dx\\ & ={(-\frac{3}{x}-\ln |x+2\left|\right)|}_2^4\\ & =-\frac{3}{4}-\ln 6+\frac{3}{2}+\ln 4=\frac{3}{4}+\ln \frac{2}{3}\end{array}\end{array}$

${\int }_{-1/3}^{1/3}\frac{9x^2+4}{9x^2-4}dx$ (20)

$\begin{array}{rl}& \frac{9x^2+4}{9x^2-4}=1+\frac{8}{9x^2-4}\\ & \frac{8}{9x^2-4}=\frac{8}{(3x-2)(3x+2)}=\frac{A}{3x-2}+\frac{B}{3x+2}\\ & \Rightarrow 8=A(3x+2)+B(3x-2)\\ & x=\frac{2}{3}\Rightarrow A=2\\ & x=-\frac{2}{3}\Rightarrow B=-2\\ & {\int }_{-\frac{1}{3}}^{\frac{1}{3}}\frac{9x^2+4}{9x^2-4}dx={\int }_{-\frac{1}{3}}^{\frac{1}{3}}(1+\frac{2}{3x-2}+\frac{-2}{3x+2})dx\\ & ={(x+\frac{2}{3}\ln |3x-2|-\frac{2}{3}\ln |3x+2\left|\right)|}_{-\frac{1}{3}}^{\frac{1}{3}}\\ & ={(x+\frac{2}{3}\ln \left|\frac{3x-2}{3x+2}\right|)|}_{-\frac{1}{3}}^{\frac{1}{3}}\\ & =\frac{1}{3}+\frac{2}{3}\ln \frac{1}{3}+\frac{1}{3}-\frac{2}{3}\ln 3=\frac{2}{3}-\frac{4}{3}\ln 3\\ & \end{array}$

${\int }_0^1\frac{17-5x}{(2x+3)(2-x{)}^2}dx$ (21)

$\begin{array}{rl}& \frac{17-5x}{(2x+3)(2-x{)}^2}=\frac{A}{2x+3}+\frac{B}{2-x}+\frac{C}{(2-x{)}^2}\\ & \Rightarrow 17-5x=A(2-x{)}^2+B(2-x\left)\right(2x+3)+C(2x+3)\\ & x=-\frac{3}{2}\Rightarrow A=2\\ & x=2\Rightarrow C=1\\ & x=0\Rightarrow 17=4A+6B+3C\Rightarrow B=1\\ & {\int }_0^1\frac{17-5x}{(2x+3)(2-x{)}^2}dx={\int }_0^1(\frac{2}{2x+3}+\frac{1}{2-x}+\frac{1}{(2-x{)}^2})dx\\ & ={(ln|2x+3|-ln|2-x|+\frac{1}{2-x})|}_0^1\\ & =ln5+1-ln3+ln2-\frac{1}{2}=\frac{1}{2}+ln\frac{10}{3}\end{array}$

${\int }_1^4\frac{4}{16x^2+8x-3}dx$ (22)

$\begin{array}{rl}& \frac{4}{16x^2+8x-3}=\frac{4}{(4x-1)(4x+3)}=\frac{A}{4x-1}+\frac{B}{4x+3}\\ & \Rightarrow 4=A(4x+3)+B(4x-1)\\ & x=\frac{1}{4}\Rightarrow A=1\\ & x=-\frac{3}{4}\Rightarrow B=-1\\ & {\int }_1^4\frac{4}{16x^2+8x-3}dx={\int }_1^4(\frac{1}{4x-1}+\frac{-1}{4x+3})dx\\ & ={(\frac{1}{4}\ln |4x-1|-\frac{1}{4}\ln |4x+3\left|\right)|}_1^4\\ & ={(\frac{1}{4}\ln \left|\frac{4x-1}{4x+3}\right|)|}_1^4\\ & =\frac{1}{4}(\ln \frac{15}{19}-\ln \frac{3}{7})=\frac{1}{4}\ln \frac{35}{19}\end{array}$

${\int }_3^4\frac{5x+5}{x^2+x-6}dx$ (23)

$\begin{array}{rl}& \frac{5x+5}{x^2+x-6}=\frac{5x+5}{(x-2)(x+3)}=\frac{A}{x-2}+\frac{B}{x+3}\\ & \Rightarrow 5x+5=A(x+3)+B(x-2)\\ & x=2\Rightarrow A=3\\ & x=-3\Rightarrow B=2\\ & {\int }_3^4\frac{5x+5}{x^2+x-6}dx={\int }_3^4(\frac{3}{x-2}+\frac{2}{x+3})dx\\ & ={(3\ln |x-2|+2\ln |x+3\left|\right)|}_3^4\\ & =3\ln 2+2\ln 7-2\ln 6=\ln \frac{98}{9}\end{array}$

${\int }_3^4\frac{4}{x^3-4x^2+4x}dx$ (24)

$bbb\begin{array}{rl}& \frac{4}{x^3-4x^2+4x}=\frac{4}{x(x-2{)}^2}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2{)}^2}\\ & \Rightarrow 4=A(x-2{)}^2+Bx(x-2)+Cx\\ & x=0\Rightarrow A=1\\ & x=2\Rightarrow C=2\\ & x=1\Rightarrow 4=A-B+C\Rightarrow B=-1\\ & A={\int }_3^4\frac{4}{x^3-4x^2+4x}dx={\int }_3^4(\frac{1}{x}+\frac{-1}{x-2}+\frac{2}{(x-2{)}^2})dx\\ & ={(\ln |x|-\ln |x-2|-\frac{2}{x-2})|}_3^4\\ & ={(\ln \left|\frac{x}{x-2}\right|-\frac{2}{x-2})|}_3^4\\ & =\ln 2-1-\ln 3+2=1+\ln \frac{2}{3}\end{array}$

أجد مساحة المنطقة المظللة في كل من التمثيلين البيانيين الآتيين:

$\begin{array}{rl}& A={\int }_0^1\frac{1}{x^2-5x+6}dx\\ & \frac{1}{x^2-5x+6}=\frac{1}{(x-3)(x-2)}=\frac{A}{x-3}+\frac{B}{x-2}\\ & \Rightarrow \mathbf{1}=A(x-2)+B(x-3)\\ & x=3\Rightarrow A=1\\ & x=2\Rightarrow B=-1\\ & A={\int }_0^1\frac{1}{x^2-5x+6}dx={\int }_0^1(\frac{1}{x-3}+\frac{-1}{x-2})dx\\ & ={(\ln |x-3|-\ln |x-2\left|\right)|}_0^1\\ & ={\ln \left|\frac{x-3}{x-2}\right||}_0^1\\ & =\ln 2-\ln \frac{3}{2}=\ln \frac{4}{3}\end{array}$

$\begin{array}{rl}& A={\int }_1^2\frac{x^2+1}{3x-x^2}dx\\ & \frac{x^2+1}{3x-x^2}=-1+\frac{3x+1}{3x-x^2}\\ & \frac{3x+1}{3x-x^2}=\frac{3x+1}{x(3-x)}=\frac{A}{x}+\frac{B}{3-x}\\ & \Rightarrow 3x+1=A(3-x)+Bx\\ & x=0\Rightarrow A=\frac{1}{3}=-2+\frac{1}{3}\ln 2+1+\frac{10}{3}\ln 2\\ & x=3\Rightarrow B=-\frac{11}{3}\ln 2\\ & {\int }_1^2\frac{x^2+1}{3x-x^2}dx={\int }_1^2(-1+\frac{10}{3}+\frac{1}{3-x})dx\end{array}$

يبين الشكل المجاور جزءاً من منحنى الاقتران: $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{5}}{\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{5}\mathbf{x}\mathbf{-}\mathbf{3}}$:

(27) أجد إحداثيي النقطة $A$.

$f\left(x\right)=0\Rightarrow 4x-5=0\Rightarrow x=\frac{5}{4}\Rightarrow A(\frac{5}{4},0)$

(28) أجد مساحة المنطقة المظللة.

$A={\int }_0^{\frac{5}{4}}\frac{4x-5}{2x^2-5x-3}dx=\ln {|2x^2-5x-3|}_0^{\frac{5}{4}}=\ln \frac{49}{8}-\ln 3=\ln \frac{49}{24}$

ملاحظة: البسط هو مشتقة المقام، فلا داعي لتجزئة الكسر.

أجد كلاً من التكاملات الآتية:

$\int \frac{\sin x}{\cos x+{\cos }^{2}x}dx$ (29)

$\begin{array}{rl}& u=\cos x\Rightarrow \frac{du}{dx}=-\sin x\Rightarrow dx=\frac{du}{-\sin x}\\ & \int \frac{\sin x}{\cos x+{\cos }^{2}x}dx=\int \frac{\sin x}{u+u^2}\times \frac{du}{-\sin x}=\int \frac{-1}{u+u^2}du\\ & \frac{-1}{u+u^2}=\frac{-1}{u(1+u)}=\frac{A}{u}+\frac{B}{1+u}\\ & \Rightarrow -1=A(1+u)+Bu\\ & u=0\Rightarrow A=-1\\ & u=-1\Rightarrow B=1\\ & \int \frac{-1}{u+u^2}du=\int (\frac{-1}{u}+\frac{1}{1+u})du\\ & \Rightarrow \int \frac{\sin x}{\cos x+{\cos }^{2}x}dx=-\ln |\cos x|+\ln |1+\cos x|+C\\ & =\ln |1+u|+C\\ & \frac{1+\cos x}{\cos x}|+C=\ln |1+\sec x\mid +C\end{array}$

$\int \frac{1}{x^2+x\sqrt{x}}dx$ (30)

$\begin{array}{rl}& u=\sqrt{x}\Rightarrow u^2=x\Rightarrow dx=2udu\\ & \int \frac{1}{x^2+x\sqrt{x}}dx=\int \frac{1}{u^4+u^3}2udu=\int \frac{2}{u^3+u^2}du\\ & \frac{2}{u^3+u^2}=\frac{2}{u^2(u+1)}=\frac{A}{u}+\frac{B}{u^2}+\frac{C}{u+1}\\ & \Rightarrow 2=Au(u+1)+B(u+1)+C{u}^2\\ & u=0\Rightarrow B=2\\ & u=-1\Rightarrow C=2\\ & u=1\Rightarrow 2=2A+2B+C\Rightarrow A=-2\\ & \int \frac{2}{u^3+u^2}du=\int (\frac{-2}{u}+\frac{2}{u^2}+\frac{2}{u+1})du\\ & =-2\ln \left|u\right|-\frac{2}{u}+2\ln |u+1|+C\\ & \Rightarrow \int \frac{1}{x^2+x\sqrt{x}}dx=2\ln \left|\frac{u+1}{u}\right|-\frac{2}{u}+C\end{array}$

$\int \frac{e^{2x}}{e^{2x}+3e^x+2}dx$ (31)

$\begin{array}{rl}& u=e^x\Rightarrow \frac{du}{dx}=e^x=u⟹dx=\frac{du}{u}\\ & \int \frac{e^{2x}}{e^{2x}+3e^x+2}dx=\int \frac{u^2}{u^2+3u+2}\times \frac{du}{u}=\int \frac{u}{u^2+3u+2}du\\ & \frac{u}{u^2+3u+2}=\frac{u}{(u+1)(u+2)}=\frac{A}{u+1}+\frac{B}{u+2}\\ & \Rightarrow u=A(u+2)+B(u+1)\\ & u=-1\Rightarrow A=-1\\ & u=-2\Rightarrow B=2\\ & \int \frac{u}{u^2+3u+2}du=\int (\frac{-1}{u+1}+\frac{2}{u+2})du\\ & =-\ln |u+1|+2\ln |u+2|+C\\ & \Rightarrow \int \frac{e^{2x}}{e^{2x}+3e^x+2}dx=-\ln (e^x+1)+2\ln (e^x+2)+C\end{array}$

$\int \frac{\cos x}{\sin x({\sin }^{2}x-4)}dx$ (32)

$\begin{array}{rl}& u=\sin x\Rightarrow \frac{du}{dx}=\cos x\Rightarrow dx=\frac{du}{\cos x}\\ & \int \frac{\cos x}{\sin x({\sin }^{2}x-4)}dx=\int \frac{\cos x}{u(u^2-4)}\times \frac{du}{\cos x}=\int \frac{1}{u(u^2-4)}du\\ & \frac{1}{u(u^2-4)}=\frac{1}{u(u-2)(u+2)}=\frac{A}{u}+\frac{B}{u-2}+\frac{C}{u+2}\\ & \Rightarrow 1=A(u-2)(u+2)+Bu(u+2)+Cu(u-2)\\ & u=0\Rightarrow A=-\frac{1}{4}\\ & u=2\Rightarrow B=\frac{1}{8}\\ & u=-2\Rightarrow C=\frac{1}{8}\\ & \int \frac{1}{u(u^2-4)}du=\int (\frac{-\frac{1}{4}}{u}+\frac{\frac{1}{8}}{u-2}+\frac{\frac{1}{8}}{u+2})du\\ & =-\frac{1}{4}\ln \left|u\right|+\frac{1}{8}\ln |u-2|+\frac{1}{8}\ln |u+2|+C\\ & \Rightarrow \int \frac{\cos x}{\sin x({\sin }^{2}x-4)}dx\\ & =-\frac{1}{4}\ln |\sin x|+\frac{1}{8}\ln |\sin x-2|+\frac{1}{8}\ln |\sin x+2|+C\end{array}$