مهارات التفكير العليا

التكامل بالكسور الجزئية

تبرير: أحل السؤالين الآتيين تباعاً:

(33) أجد: $\int \frac{dx}{1+e^x}$ بطريقيتين مختلفتين، إحداهما الكسور الجزئية، مبرراً أجايتي.

الحل الأول بضرب كل من البسط والمقام بـ $e^{-x}$

$\int \frac{1}{1+e^x}dx=\int \frac{e^{-x}}{e^{-x}+1}dx=-\int \frac{-e^{-x}}{e^{-x}+1}dx=-\ln (e^{-x}+1)+C$

الحل الثاني بالتعويض:

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

(34) أجد: ${\int }_0^{\ln 2}\frac{1}{1+e^x}dx$. 

$\begin{array}{rl}{\int }_0^{\ln 2}\frac{1}{1+e^x}dx& =\ln e^x-\ln (e^x+1){\ln }_0^{\ln 2}\\ & =\ln e^{\ln 2}-\ln (e^{\ln 2}+1)-(\ln e^0-\ln (e^0+1))\\ & =\ln 2-\ln 3-0+\ln 2=\ln 4-\ln 3=\ln \frac{4}{3}\end{array}$

(35) تبرير: أثبت أن: ${\int }_4^9\frac{5x^2-8x+1}{2x(x-1{)}^2}dx=\ln \left(\frac{32}{3}\right)-\frac{5}{24}$.

$\begin{array}{rl}& \frac{5x^2-8x+1}{2x(x-1{)}^2}=\frac{A}{2x}+\frac{B}{x-1}+\frac{C}{(x-1{)}^2}\\ & \Rightarrow 5x^2-8x+1=A(x-1{)}^2+B(2x\left)\right(x-1)+C(2x)\\ & x=0\Rightarrow A=1\\ & x=1\Rightarrow C=-1\\ & x=-1\Rightarrow 14=4A+4B-2C\Rightarrow B=2\\ & {\int }_4^9\frac{5x^2-8x+1}{2x(x-1{)}^2}dx={\int }_4^9(\frac{1}{2x}+\frac{2}{x-1}+\frac{-1}{(x-1{)}^2})dx\\ & ={(\frac{1}{2}\ln |x|+2\ln |x-1|+\frac{1}{x-1})|}_4^9\\ & =\frac{1}{2}\ln 9+2\ln 8+\frac{1}{8}-\frac{1}{2}\ln 4-2\ln 3-\frac{1}{3}\\ & =\ln 3+\ln 64+\frac{1}{8}-\ln 2-\ln 9-\frac{1}{3}\\ & =\ln \frac{3\left(64\right)}{2\left(9\right)}-\frac{5}{24}=\ln \frac{32}{3}-\frac{5}{24}\end{array}$

(36) تبرير: أثبت أن: ${\int }_9^{16}\frac{2\sqrt{x}}{x-4}dx=4(1+\ln \left(\frac{5}{3}\right))$.

$\begin{array}{rl}& u=\sqrt{x}\Rightarrow u^2=x\Rightarrow dx=2udu\\ & x=9\Rightarrow u=3\\ & x=16\Rightarrow u=4\\ & {\int }_9^{16}\frac{2\sqrt{x}}{x-4}dx={\int }_3^4\frac{2u}{u^2-4}2udu={\int }_3^4\frac{4u^2}{u^2-4}du\\ & ={\int }_3^4(4+\frac{16}{u^2-4})du\\ & \frac{16}{u^2-4}=\frac{16}{(u-2)(u+2)}=\frac{A}{u-2}+\frac{B}{u+2}\\ & \Rightarrow 16=A(u+2)+B(u-2)\\ & u=2\Rightarrow A=4\\ & u=-2\Rightarrow B=-4\\ & {\int }_3^4(4+\frac{16}{u^2-4})du={\int }_3^4(4+\frac{4}{u-2}+\frac{-4}{u+2})du\\ & ={(4u+4\ln |u-2|-4\ln |u+2\left|\right)|}_3^4\\ & =16+4\ln 2-4\ln 6-12-4\ln 1+4\ln 5\\ & =4+4\ln \frac{5}{3}=4(1+\ln \frac{5}{3})\\ & \Rightarrow {\int }_9^{16}\frac{2\sqrt{x}}{x-4}dx=4(1+\ln \frac{5}{3})\end{array}$

(37) تبرير: أثبت أن: ${\int }_0^1\frac{4x^2+9x+4}{2x^2+5x+3}dx=2+\frac{1}{2}\ln \frac{5}{12}$.

$\begin{array}{rl}& \frac{4x^2+9x+4}{2x^2+5x+3}=2-\frac{x+2}{2x^2+5x+3}\\ & \frac{x+2}{2x^2+5x+3}=\frac{x+2}{(x+1)(2x+3)}=\frac{A}{x+1}+\frac{B}{2x+3}\\ & \Rightarrow x+2=A(2x+3)+B(x+1)\\ & x=-1\Rightarrow A=1\\ & x=-\frac{3}{2}\Rightarrow B=-1\\ & {\int }_0^1\frac{4x^2+9x+4}{2x^2+5x+3}dx={\int }_0^1(2-\frac{1}{x+1}+\frac{1}{2x+3})dx\\ & ={(2x-\ln |x+1|+\frac{1}{2}\ln |2x+3\left|\right)|}_0^1\\ & =2-\ln 2+\frac{1}{2}\ln 5-0+\ln 1-\frac{1}{2}\ln 3\\ & =2-\frac{1}{2}\ln 4+\frac{1}{2}\ln 5-\frac{1}{2}\ln 3\\ & =2+\frac{1}{2}(\ln 5-\ln 4-\ln 3)=2+\frac{1}{2}\ln \frac{5}{12}\end{array}$

تحد: أجد كلاً من التكاملات الآتية:

$\int \frac{\sqrt{1+\sqrt{x}}}{x}dx$ (38)

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

$\int \frac{x}{16x^4-1}dx$ (39)

$\begin{array}{rl}& \frac{x}{16x^4-1}=\frac{x}{(4x^2+1)(2x-1)(2x+1)}=\frac{Ax+B}{4x^2+1}+\frac{C}{2x-1}+\frac{D}{2x+1}\\ & \Rightarrow x=(Ax+B)(2x-1)(2x+1)+C(4x^2+1)(2x+1)+D(4x^2+1)(2x-1)\\ & x=\frac{1}{2}\Rightarrow C=\frac{1}{8}\\ & x=-\frac{1}{2}\Rightarrow D=\frac{1}{8}\\ & x=0\Rightarrow 0=-B+C-D\Rightarrow B=0\\ & x=1\Rightarrow 1=3A+3B+15C+5D\Rightarrow A=-\frac{1}{2}\\ & \int \frac{x}{16x^4-1}dx=\int (\frac{-\frac{1}{2}x}{4x^2+1}+\frac{\frac{1}{8}}{2x-1}+\frac{\frac{1}{8}}{2x+1})dx\\ & =-\frac{1}{16}ln(4x^2+1)+\frac{1}{16}ln|2x-1|+\frac{1}{16}ln|2x+1|+C\\ & =\frac{1}{16}ln\left|\frac{4x^2-1}{4x^2+1}\right|+C\end{array}$

$\int \frac{1}{\sqrt{x}-\sqrt[3]{x}}dx$ (40)

$\begin{array}{rl}& u=x^{\frac{1}{6}}\Rightarrow \frac{du}{dx}\Rightarrow \frac{1}{6}{x}^{\frac{-5}{6}}\Rightarrow dx=6x^{\frac{5}{6}}du=6u^{5}du\\ & u=x^{\frac{1}{6}}\Rightarrow x=u^6\Rightarrow \sqrt{x}=u^3,\sqrt[3]{x}=u^2\\ & \Rightarrow \int \frac{1}{\sqrt{x}-\sqrt[3]{x}}dx=\int \frac{6u^5}{u^3-u^2}du\\ & =\int \frac{6u^3}{u-1}du\\ & =\int (6u^2+6u+6+\frac{6}{u-1})du\\ & =2u^3+3u^2+6u+6\ln |u-1|+C\\ & =2\sqrt{x}+3\sqrt[3]{x}+6\sqrt[6]{x}+6\ln |\sqrt[6]{x}-1|+C\end{array}$