تمارين (1-4)

(1)- جد تكاملات كلاً مما يأتي:

$\int (6x^2+4x+3)dx$

$\begin{array}{rl}& \int (6x^2+4x+3)dx\\ & =\frac{6x^3}{3}+\frac{4x^2}{2}+3x+C=2x^3+2x^2+3x+C\end{array}$

$\int (3x-1)(x+5)dx$

$\begin{array}{rl}& \int (3x-1)(x+5)dx\\ & =\int (3x^2+15x-x-5)dx\\ & =\int (3x^2+14x-5)dx\\ & =\frac{3x^3}{3}+\frac{14x^3}{2}-5x+C=x^3+7x^2-5x+C\end{array}$

$\int \sqrt{x}(\sqrt{x}+1{)}^{2}dx$

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

$\int \frac{x^3+27}{x+3}dx$

$$\begin{gathered} \int \frac{x^3+27}{x+3}dx \\ \begin{array}{rl}& =\int \frac{(x+3)(x^2-3x+9)}{x+3}dx\\ & =\int (x^2-3x+9)dx\\ & =\frac{x^3}{3}-\frac{3x^2}{2}+9x+C\end{array} \end{gathered}$$

$\int \frac{x^3-2x^2+1}{5x^5}dx$

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

$\int \frac{x^2+2}{\sqrt[3]{x^3+6x+1}}dx$

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

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

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

$\int \frac{dx}{\sqrt[5]{x^2+16x+64}}$

$$\begin{gathered} \int \frac{dx}{\sqrt[5]{x^2+16x+64}} \\ \begin{array}{rl}=& \int \frac{dx}{\sqrt[5]{(x+8{)}^2}}=\int \frac{dx}{(x+8{)}^{\frac{2}{5}}}\Rightarrow \int (x+8{)}^{\frac{-2}{5}}dx=\frac{(x+8{)}^{\frac{3}{5}}}{\frac{3}{5}}+C\\ & \frac{5}{3}(x+8{)}^{\frac{3}{5}}+C=\frac{5}{3}\sqrt[5]{(x+8{)}^3}+C\end{array} \end{gathered}$$

$\int \sqrt[7]{2x^9-3x^7}dx$

ملاحظة $\sqrt[7]{x^7}=x\Leftarrow$

$\begin{array}{rl}& \int \sqrt[7]{x^7(2x^2-3)}dx\\ & \int {(2x^2-3)}^{\frac{1}{7}}\times dx\overline{f}\left(x\right)=4x\\ & \frac{1}{4}\int {(2x^2-3)}^{\frac{1}{7}}4xdx\Rightarrow \frac{1}{4}\cdot \frac{{(2x^2-3)}^{\frac{8}{7}}}{\frac{8}{7}}+C\Rightarrow \frac{1}{4}\cdot \frac{7}{8}{(2x^2-3)}^{\frac{8}{7}}+C\\ & =\frac{7}{32}{(2x^2-3)}^{\frac{8}{7}}+C=\frac{7}{32}\sqrt[7]{{(2x^2-3)}^8}+C\end{array}$

$\int (3x^2+\frac{1}{\sqrt{x}})dx$

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

$\int \frac{ydy}{{(19-2y^2)}^{\frac{1}{3}}}$

$\begin{array}{rl}& \int {(19-2y^2)}^{\frac{-1}{3}}ydy\\ & \overline{y}=-4y\\ & -\frac{1}{4}\int {(19-2y^2)}^{\frac{-1}{3}}(-4ydy)=\frac{-1}{4}\cdot \frac{{(19-2y^2)}^{\frac{2}{3}}}{\frac{2}{3}}+C=\frac{-1}{4}\cdot \frac{3}{2}{(19-2y^2)}^{\frac{2}{3}}+C=\frac{-3}{8}{(19-2y^2)}^{\frac{2}{3}}+C\end{array}$

$\int \frac{x^4-16}{x+2}dx$

$$\begin{gathered} \int \frac{x^4-16}{x+2}dx \\ \begin{array}{rl}& =\int \frac{(x^2-4)(x^2+4)}{x+2}dx=\int \frac{(x-2)(x+2)(x^2+4)}{(x+2)}dx=\int (x-2)(x^2+4)dx\\ & =\int (x^3+4x-2x^2-8)dx=\frac{x^4}{4}+\frac{4x^2}{2}-\frac{2x^2}{3}-8x+C=\frac{x4}{4}+2x^2-\frac{2x^3}{3}-8x+C\end{array} \end{gathered}$$

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

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