تمارين (3-4)

1- جد تكاملات كلاً مما يأتي:

- $\begin{array}{rl}& \int (2x+5)(x+1)dx\end{array}$

$\begin{array}{rl}& \int (2x+5)(x+1)dx\\ & =\int (2x^2+2x+5x+5)dx=\int (2x^2+7x+5)dx=\frac{2x^3}{3}+\frac{7x^2}{2}+5x+C\end{array}$

- $\begin{array}{rl}& {\int }_{-1}^1(x^2+3)(x-2)dx\end{array}$

$\begin{array}{rl}& {\int }_{-1}^1(x^2+3)(x-2)dx\\ & ={\int }_{-1}^1(x^3-2x^2+3x-6)dx={[\frac{x^4}{4}-\frac{2x^3}{3}+\frac{3x^2}{2}-6x]}_{-1}^{-1}=(\frac{1}{4}-\frac{2}{3}+\frac{3}{2}-6)-(\frac{1}{4}+\frac{2}{3}+\frac{3}{2}+6)\\ & =\frac{1}{4}-\frac{2}{3}+\frac{3}{2}-6-\frac{1}{4}-\frac{2}{3}-\frac{3}{2}-6)=-\frac{2}{3}-6-\frac{2}{3}-6=-\frac{4}{3}-12=\frac{-4-36}{3}=\frac{-40}{3}\end{array}$

- $\begin{array}{rl}& \int \sqrt{x}(\sqrt{x}+5)dx\end{array}$

$\begin{array}{rl}& \int \sqrt{x}(\sqrt{x}+5)dx\\ & =\int x^{\frac{1}{2}}(x^{\frac{1}{2}}+5)dx=\int (x+5x^{\frac{1}{2}})dx=\int \frac{x^2}{2}+\frac{5x^{\frac{3}{2}}}{\frac{3}{2}}+c=\frac{x^2}{2}+5\left(\frac{2}{3}\right)x^{\frac{3}{2}}+C=\frac{x^2}{2}+\frac{10}{3}{x}^{\frac{3}{2}}+C\end{array}$

- $\begin{array}{rl}& {\int }_0^4\sqrt{x}(x+1{)}^{2}dx\end{array}$

$\begin{array}{rl}& {\int }_0^4\sqrt{x}(x+1{)}^{2}dx\\ =& {\int }_0^4{x}^{\frac{1}{2}}(x^2+2x+1)dx={\int }_0^4(x^{\frac{5}{2}}+2x^{\frac{3}{2}}+x^{\frac{1}{2}})dx={[\frac{2}{7}{x}^{\frac{7}{2}}+2\frac{2}{5}{x}^{\frac{5}{2}}+\frac{2}{3}{x}^{\frac{3}{2}}]}_0^4={[\frac{2}{7}{x}^{\frac{7}{2}}+\frac{4}{5}{x}^{\frac{5}{2}}+\frac{2}{3}{x}^{\frac{3}{2}}]}_0^4\\ =& {[\frac{2}{7}\sqrt{x^7}+\frac{4}{5}\sqrt{x^5}+\frac{2}{3}\sqrt{x^3}]}_0^4=(\frac{2}{7}\sqrt{(4{)}^7}+\frac{4}{5}\sqrt{(4{)}^5}+\frac{2}{3}\sqrt{(4{)}^3})-\left(0\right)\\ =& \left(\frac{2}{7}\right(2{)}^7+\frac{4}{5}(2{)}^5+\frac{2}{3}(2{)}^3)=\left(\frac{2}{7}\right(128)+\frac{4}{5}(32)+\frac{2}{3}(8\left)\right)\\ =& \frac{256}{7}+\frac{128}{5}+\frac{16}{3}=\frac{3840+2688+560}{105}=\frac{7088}{105}\end{array}$

- $\begin{array}{rl}& \int \sqrt{x}(\sqrt{x}+2{)}^{2}dx\end{array}$

$\begin{array}{rl}& \int \sqrt{x}(\sqrt{x}+2{)}^{2}dx\\ =& \int x^{\frac{1}{2}}(x+2x^{\frac{1}{2}}+4)dx=\int (x^{\frac{3}{2}}+4x+4x^{\frac{1}{2}})dx=\frac{2}{5}{x}^{\frac{5}{2}}+2x^2+\frac{8}{3}{x}^{\frac{3}{2}}+C\end{array}$

- ${\int }_{-1}^0\frac{x^3-27}{x-3}dx$

$$\begin{gathered} {\int }_{-1}^0\frac{x^3-27}{x-3}dx \\ \begin{array}{rl}& ={\int }_{-1}^0\frac{(x-3)(x^2+3x+9)}{x-3}dx={[\frac{x^3}{3}+\frac{3x^2}{2}+9x]}_{-1}^0=\left(0\right)-(-\frac{1}{3}+\frac{3}{2}-9)=\frac{1}{3}-\frac{3}{2}+9\\ & =\frac{2-9+54}{6}=\frac{-7+54}{6}=\frac{47}{6}\end{array} \end{gathered}$$

- $\begin{array}{rl}& \int \frac{x^4-1}{x-1}dx\end{array}$

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

- ${\int }_0^1\frac{xdx}{\sqrt{x^2+1}}$

$\begin{array}{rl}& {\int }_0^1\frac{xdx}{\sqrt{x^2+1}}\Rightarrow {\int }_0^1\frac{xdx}{{(x^2+1)}^{\frac{1}{2}}}\\ & ={\int }_0^1{(x^2+1)}^{-\frac{1}{2}}xdx,\overline{f}\left(x\right)=2x\\ & =\frac{1}{2}{\int }_0^1{(x^2+1)}^{\frac{-1}{2}}xdx=\frac{1}{2}{\left[\frac{{(x^2+1)}^{\frac{1}{2}}}{\frac{1}{2}}\right]}_0^1={\left[\sqrt{x^2+1}\right]}_0^1=(\sqrt{1+1}-(\sqrt{0+1})=\sqrt{2}-1\end{array}$

- $\int \frac{x^2+1dx}{\sqrt[3]{x^3+3x+1}}$

$$\begin{gathered} \int \frac{x^2+1dx}{\sqrt[3]{x^3+3x+1}} \\ \begin{array}{rl}& =\int \frac{x^2+1dx}{{(x^3+3x+1)}^{\frac{1}{3}}}=\int {(x^3+3x+1)}^{\frac{-1}{3}}(x^2+1)dx\\ & =\frac{1}{3}\int {(x^3+3x+1)}^{\frac{-1}{3}}(3x^2+3)dx=\frac{1}{3}\frac{{(x^3+3x+1)}^{\frac{2}{3}}}{\frac{2}{3}}+C\overline{f}\left(x\right)=3x^2+3\\ & =\frac{1}{3}\cdot \frac{3}{2}{(x^3+3x+1)}^{\frac{2}{3}}+C=\frac{1}{2}{(x^3+3x+1)}^{\frac{2}{3}}+C\end{array} \end{gathered}$$

- ${\int }_0^3\sqrt[3]{(3x-1{)}^2}dx$

$$\begin{gathered} {\int }_0^3\sqrt[3]{(3x-1{)}^2}dx \\ \begin{array}{rl}& ={\int }_0(3x-1{)}^{\frac{2}{3}}dx\overline{f}(x)=3\\ & =\frac{1}{3}{\int }_0^3(3x-1{)}^{\frac{2}{3}}3dx=\frac{1}{3}{\left[\frac{(3x-1{)}^{\frac{5}{3}}}{\frac{5}{3}}\right]}_0^3\\ & =\frac{1}{3}\cdot \frac{3}{5}{\left[\right(3x-1{)}^{\frac{2}{3}}]}_0^3=\frac{1}{5}{\left[\sqrt[3]{(3x-1{)}^5}\right]}_0^3=\frac{1}{5}[\sqrt[3]{(8{)}^5}-\sqrt[3]{(0-1{)}^5}\\ & =\frac{1}{5}\left[\right(2{)}^2+1]=\frac{1}{5}[32+1]=\frac{33}{5}\end{array} \end{gathered}$$

- $\int \frac{\sqrt[3]{x}+1}{\sqrt[3]{x^2}}dx$

$$\begin{gathered} \int \frac{\sqrt[3]{x}+1}{\sqrt[3]{x^2}}dx \\ =\int \frac{x^{\frac{1}{3}}+1}{x^{\frac{2}{3}}}dx=\int x^{\frac{-2}{3}}(x^{\frac{1}{3}}+1)dx=\int (x^{-\frac{1}{3}}+x^{-\frac{2}{3}})dx=\frac{3}{2}{x}^{\frac{2}{3}}+3x^{\frac{1}{3}}+C \end{gathered}$$

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

$$\begin{gathered} \int \frac{\sqrt[3]{\sqrt{x}-1}}{\sqrt{x}}dx \\ \begin{array}{rl}& =\int \frac{{(x^{\frac{1}{2}}-1)}^{\frac{1}{3}}}{x^{\frac{1}{2}}}dx\Rightarrow {(x^{\frac{1}{2}}-1)}^{\frac{1}{3}}{x}^{\frac{-1}{2}}dx\\ & =2\int {(x^{\frac{1}{2}}-1)}^{\frac{1}{3}}\frac{1}{2}{x}^{-\frac{1}{2}}dx=2\frac{(\sqrt{x}-1{)}^{\frac{4}{3}}}{\frac{4}{3}}+C=2\frac{3}{4}(\sqrt{x}-1{)}^{\frac{4}{3}}+C=\frac{3}{2}(\sqrt{x}-1{)}^{\frac{4}{3}}+C\end{array} \end{gathered}$$

- $\int \frac{x^4}{\sqrt[5]{a^2{x}^5+b^2}}dx$

$$\begin{gathered} \int \frac{x^4}{\sqrt[5]{a^2{x}^5+b^2}}dx \\ \begin{array}{rl}& =\int {(a^2{x}^5+b^2)}^{-\frac{1}{5}}{x}^{3}dx,\overline{f}\left(x\right)=5a^2{x}^4\\ & =\frac{1}{5a^2}\int {(a^2{x}^5+b^2)}^{-\frac{1}{5}}5a^2{x}^{4}dx=\frac{1}{5a^2}\frac{{(a^2{x}^5+b^2)}^{\frac{4}{5}}}{\frac{4}{5}}+C=\frac{1}{5a^2}\cdot \frac{5}{4}{(a^2{x}^5+b^2)}^{\frac{4}{5}}+C\\ & =\frac{1}{4a^2}{(a^2{x}^5+b^2)}^{\frac{4}{5}}+C\end{array} \end{gathered}$$

- ${\int }_0^8\sqrt{x^2-14x+49}dx$

ملاحظة: ${\int }_a^b\sqrt{x^2}dx{\int }_a^b\left|x\right|dx$

$\begin{array}{rl}& {\int }_0^8(x-7{)}^{2}dx={\int }_0^8|x-7|dx\\ & |x-7|=\{\begin{array}{c}x-7,x\ge 7\\ -x+7,x<7\\ 8\end{array}\\ & {\int }_0^7(-x+7)dx+{\int }_7^7(x-7)dx={[\frac{-x^2}{2}+7x]}_0^7+{[\frac{x^2}{2}-7x]}_7^8\\ & =[(\frac{-49}{2}+49)-(0\left)\right]+[(\frac{64}{2}-56)-(\frac{49}{2}-49)]=\left[\frac{-49+98}{2}\right]+\left[\right(32-56)-\left(\frac{49-98}{2}\right)]\\ & =\left[\frac{49}{2}\right]+\left[\right(-24)-\left(\frac{-49}{2}\right)]=\left(\frac{49}{2}\right)+[-24+\frac{49}{2}]=\left(\frac{49}{2}\right)+\left(\frac{-48+49}{2}\right)=\frac{49}{2}+\frac{1}{2}=\frac{50}{2}=25\end{array}$

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

$\begin{array}{rl}& \Rightarrow \int (2x-3{)}^{-2}dx\overline{f}(x)=2\\ & =\frac{1}{2}\int (2x-3{)}^{-2}2dx=\frac{1}{2}\frac{(2x-3{)}^{-1}}{-1}+C=\frac{-1}{2(2x-3)}+C\end{array}$

- ${\int }_{-1}^1\sqrt[5]{3x^5-2x^7}dx$

$$\begin{gathered} {\int }_{-1}^1\sqrt[5]{3x^5-2x^7}dx \\ \begin{array}{rl}& ={\int }_{-1}^1\sqrt[5]{x^5(3-2x^2)}dx={\int }_{-1}^1{(3-2x^2)}^{\frac{1}{5}}xdx\overline{f}\left(x\right)=-4x\\ & =\frac{-1}{4}{\int }_{-1}^1{(3-2x^2)}^{\frac{1}{5}}-4xdx=\frac{-1}{4}{\left[\frac{{(3-2x^2)}^{\frac{6}{5}}}{\frac{6}{5}}\right]}_{-1}^1\\ & =-\frac{1}{4}\cdot \frac{5}{6}{\left[{(3-2x^2)}^{\frac{6}{5}}\right]}_{-1}^1=\frac{-5}{24}{\left[{(3-2x^2)}^{\frac{6}{5}}\right]}_{-1}^1=\frac{-5}{24}\left[\right(3-2{)}^{\frac{6}{5}}-(3-2{)}^{\frac{6}{5}}]=\frac{-5}{24}[1-1]=0\end{array} \end{gathered}$$

- $\int \sqrt[3]{2x^5-7x^3}dx$

$$\begin{gathered} \int \sqrt[3]{2x^5-7x^3}dx \\ \begin{array}{rl}& =\int \sqrt[3]{x^3(2x^2-7)}dx\\ & =\int {(2x^2-7)}^{\frac{1}{3}}\times dx,\overline{f}\left(x\right)=4x\\ & =\frac{1}{4}\int {(2x^2-7)}^{\frac{1}{3}}4xdx=\frac{1}{4}\frac{{(2x^2-7)}^{\frac{4}{3}}}{\frac{4}{3}}+C=\frac{1}{4}\frac{3}{4}{(2x^2-7)}^{\frac{4}{3}}+C=\frac{3}{16}{(2x^2-7)}^{\frac{4}{3}}+C\end{array} \end{gathered}$$

- ${\int }_1^b(13-4x)dx=9$ جد قيمة b

$\begin{array}{rl}& {[13x-2x^2]}_1^b=9\Rightarrow (13b-2b^2)-(13-2)=9\\ & =13b-2b^2-11-9=0\Rightarrow 13b-2b^2-20=0\\ & (-2b^2+13b-20=0)(-1)\Rightarrow 2b^2-13b+20=0\Rightarrow (b-4)(2b-5)=0\\ & \mathrm{إما} b=4 \mathrm{أو} b=\frac{5}{2}\end{array}$