حلول الأسئلة

السؤال

جد تكاملات كلاً مما يأتي:

الحل

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

$\int (3 x^{2} + \frac{1}{\sqrt{x}}) d x = \int (3 x^{2} + \frac{1}{x^{\frac{1}{2}}}) d x = \int (3 x^{2} + x^{- \frac{1}{2}}) d x = \frac{3 x^{3}}{3} + \frac{x^{\frac{1}{2}}}{\frac{1}{2}} + C = x^{3} + 2 x^{\frac{1}{2}} + C = x^{3} + 2 \sqrt{x} + C$

مشاركة الحل

فايسبوك

واتساب

تيليجرام

طباعة

تمارين (1-4)

(1)- جد تكاملات كلاً مما يأتي:

$\int (6 x^{2} + 4 x + 3) d x$

$\begin{aligned} \int (6 x^{2} + 4 x + 3) d x \\ = \frac{6 x^{3}}{3} + \frac{4 x^{2}}{2} + 3 x + C = 2 x^{3} + 2 x^{2} + 3 x + C \end{aligned}$

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

$\begin{aligned} \int (3 x - 1) (x + 5) d x \\ = \int (3 x^{2} + 15 x - x - 5) d x \\ = \int (3 x^{2} + 14 x - 5) d x \\ = \frac{3 x^{3}}{3} + \frac{14 x^{3}}{2} - 5 x + C = x^{3} + 7 x^{2} - 5 x + C \end{aligned}$

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

$\begin{aligned} \int \sqrt{x} (\sqrt{x} + 1)^{2} d x \\ = & \int \sqrt{x} (x + 2 \sqrt{x} + 1) d x = \int_{\frac{5}{2}} x^{\frac{1}{2}} (x + 2 x^{\frac{1}{2}} + 1) d x \\ = \int (x^{\frac{3}{2}} + 2 x + x^{\frac{1}{2}}) d x = \frac{x^{\frac{5}{2}}}{2} + \frac{2 x^{2}}{2} + \frac{x^{\frac{3}{2}}}{\frac{3}{2}} + C \\ = \frac{2}{5} x^{\frac{5}{2}} + x^{2} + \frac{2}{3} x^{\frac{3}{2}} + C \end{aligned}$

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

$\int \frac{x^{3} + 27}{x + 3} d x \begin{aligned} = \int \frac{(x + 3) (x^{2} - 3 x + 9)}{x + 3} d x \\ = \int (x^{2} - 3 x + 9) d x \\ = \frac{x^{3}}{3} - \frac{3 x^{2}}{2} + 9 x + C \end{aligned}$

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

$\int \frac{x^{3} - 2 x^{2} + 1}{5 x^{5}} d x \begin{aligned} = \frac{1}{5} \int x^{- 5} (x^{3} - 2 x^{2} + 1) d x \\ = \frac{1}{5} \int (x^{- 2} - 2 x^{- 3} + x^{5}) d x \\ = \frac{1}{5} [\frac{x^{- 1}}{- 1} - \frac{2 x^{- 2}}{- 2} + \frac{x^{- 4}}{- 4}] + C \\ = \frac{1}{5} [- x^{- 1} + x^{- 2} - \frac{x^{- 4}}{+ 4}] + C \\ = \frac{1}{5} [\frac{- 1}{x} + \frac{1}{x^{2}} - \frac{1}{4 x^{4}}] + C \end{aligned}$

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

$\int \frac{x^{2} + 2}{\sqrt[3]{x^{3} + 6 x + 1}} d x \begin{aligned} = \int \frac{x^{2} + 2}{{(x^{3} + 6 x + 1)}^{\frac{1}{3}}} d x \\ = \int_{= \frac{1}{3}} \int {(x^{3} + 6 x + 1)}^{- \frac{1}{3}} (x^{2} + 2) d x \\ = \frac{1}{3} \frac{{(x^{3} + 6 x + 1)}^{- \frac{1}{3}} (3 x^{2} + 6) d x \bar{f} (x) = 3 x^{2} + 6}{\frac{2}{3}} + C \\ = \frac{1}{3} \cdot \frac{3}{2} {(x^{3} + 6 x + 1)}^{\frac{2}{3}} + C \\ = \frac{1}{2} {(x^{2} + 6 x + 1)}^{\frac{2}{3}} + C \\ = \frac{1}{2} \sqrt[3]{{(x^{3} + 6 x + 1)}^{2}} + C \end{aligned}$

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

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

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

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

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

ملاحظة $\sqrt[7]{x^{7}} = x \Leftarrow$

$\begin{aligned} \int \sqrt[7]{x^{7} (2 x^{2} - 3)} d x \\ \int {(2 x^{2} - 3)}^{\frac{1}{7}} \times d x \bar{f} (x) = 4 x \\ \frac{1}{4} \int {(2 x^{2} - 3)}^{\frac{1}{7}} 4 x d x \Rightarrow \frac{1}{4} \cdot \frac{{(2 x^{2} - 3)}^{\frac{8}{7}}}{\frac{8}{7}} + C \Rightarrow \frac{1}{4} \cdot \frac{7}{8} {(2 x^{2} - 3)}^{\frac{8}{7}} + C \\ = \frac{7}{32} {(2 x^{2} - 3)}^{\frac{8}{7}} + C = \frac{7}{32} \sqrt[7]{{(2 x^{2} - 3)}^{8}} + C \end{aligned}$

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

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

$\begin{aligned} \int {(19 - 2 y^{2})}^{\frac{- 1}{3}} y d y \\ \bar{y} = - 4 y \\ - \frac{1}{4} \int {(19 - 2 y^{2})}^{\frac{- 1}{3}} (- 4 y d y) = \frac{- 1}{4} \cdot \frac{{(19 - 2 y^{2})}^{\frac{2}{3}}}{\frac{2}{3}} + C = \frac{- 1}{4} \cdot \frac{3}{2} {(19 - 2 y^{2})}^{\frac{2}{3}} + C = \frac{- 3}{8} {(19 - 2 y^{2})}^{\frac{2}{3}} + C \end{aligned}$

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

$\int \frac{x^{4} - 16}{x + 2} d x \begin{aligned} = \int \frac{(x^{2} - 4) (x^{2} + 4)}{x + 2} d x = \int \frac{(x - 2) (x + 2) (x^{2} + 4)}{(x + 2)} d x = \int (x - 2) (x^{2} + 4) d x \\ = \int (x^{3} + 4 x - 2 x^{2} - 8) d x = \frac{x^{4}}{4} + \frac{4 x^{2}}{2} - \frac{2 x^{2}}{3} - 8 x + C = \frac{x 4}{4} + 2 x^{2} - \frac{2 x^{3}}{3} - 8 x + C \end{aligned}$

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

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

مشاركة الدرس

فايسبوك

واتساب

تيليجرام

طباعة

السؤال

جد تكاملات كلاً مما يأتي:

الحل

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

تمارين (1-4)

(1)- جد تكاملات كلاً مما يأتي:

$\int (6 x^{2} + 4 x + 3) d x$

$\begin{aligned} \int (6 x^{2} + 4 x + 3) d x \\ = \frac{6 x^{3}}{3} + \frac{4 x^{2}}{2} + 3 x + C = 2 x^{3} + 2 x^{2} + 3 x + C \end{aligned}$

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

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

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

$\int \frac{x^{3} + 27}{x + 3} d x \begin{aligned} = \int \frac{(x + 3) (x^{2} - 3 x + 9)}{x + 3} d x \\ = \int (x^{2} - 3 x + 9) d x \\ = \frac{x^{3}}{3} - \frac{3 x^{2}}{2} + 9 x + C \end{aligned}$

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

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

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

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

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

ملاحظة $\sqrt[7]{x^{7}} = x \Leftarrow$

حلول الأسئلة

جد تكاملات كلاً مما يأتي:

مشاركة الحل

تمارين (1-4)

(1)- جد تكاملات كلاً مما يأتي:

∫(6x2+4x+3)dx

∫(3x−1)(x+5)dx

∫x(x+1)2dx

∫x3+27x+3dx

∫x3−2x2+15x5dx

∫x2+2x3+6x+13dx

∫x23+2x3dx

∫dxx2+16x+645

∫2x9−3x77dx

∫(3x2+1x)dx

∫ydy(19−2y2)13

∫x4−16x+2dx

∫(x3−1x3dx)

مشاركة الدرس

جد تكاملات كلاً مما يأتي:

تمارين (1-4)

(1)- جد تكاملات كلاً مما يأتي:

∫(6x2+4x+3)dx

∫(3x−1)(x+5)dx

∫x(x+1)2dx

∫x3+27x+3dx

∫x3−2x2+15x5dx

∫x2+2x3+6x+13dx

∫x23+2x3dx

∫dxx2+16x+645

∫2x9−3x77dx

∫(3x2+1x)dx

∫ydy(19−2y2)13

∫x4−16x+2dx

∫(x3−1x3dx)

$\int (6 x^{2} + 4 x + 3) d x$

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

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

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

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

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

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

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

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

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

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

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

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

$\int (6 x^{2} + 4 x + 3) d x$

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

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

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

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

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

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

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

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

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

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

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

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