مهارات التفكير العليا

مهارات التفكير العليا

التكامل بالكسور الجزئية

تبرير: أحل السؤالين الآتيين تباعاً:

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

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

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

الحل الثاني بالتعويض:

$\begin{aligned} u = e^{x} \Rightarrow \frac{d u}{d x} = e^{x} = u \Rightarrow d x = \frac{d u}{u} \\ \int \frac{1}{1 + e^{x}} d x = \int \frac{1}{1 + u} \times \frac{d u}{u} = \int \frac{1}{u (1 + u)} d u \\ \frac{1}{u (1 + u)} = \frac{A}{u} + \frac{B}{u + 1} \\ \Rightarrow 1 = A (1 + u) + B u \\ u = 0 \Rightarrow A = 1 \\ u = - 1 \Rightarrow B = - 1 \\ \int \frac{1}{u (1 + u)} d u = \int (\frac{1}{u} + \frac{- 1}{u + 1}) d u = l n | u | - l n | u + 1 | + C \\ \Rightarrow \int \frac{1}{1 + e^{x}} d x = l n e^{x} - l n (e^{x} + 1) + C \\ = l n {(\frac{e^{x} + 1}{e^{x}})}^{- 1} + C = - l n (e^{- x} + 1) + C \end{aligned}$

(34) أجد: $\int_{0}^{\ln 2} \frac{1}{1 + e^{x}} d x$ .

$\begin{aligned} \int_{0}^{\ln 2} \frac{1}{1 + e^{x}} d x & = \ln e^{x} - \ln (e^{x} + 1) \ln_{0}^{\ln 2} \\ = \ln e^{\ln 2} - \ln (e^{\ln 2} + 1) - (\ln e^{0} - \ln (e^{0} + 1)) \\ = \ln 2 - \ln 3 - 0 + \ln 2 = \ln 4 - \ln 3 = \ln \frac{4}{3} \end{aligned}$

(35) تبرير: أثبت أن: $\int_{4}^{9} \frac{5 x^{2} - 8 x + 1}{2 x (x - 1)^{2}} d x = \ln (\frac{32}{3}) - \frac{5}{24}$ .

$\begin{aligned} \frac{5 x^{2} - 8 x + 1}{2 x (x - 1)^{2}} = \frac{A}{2 x} + \frac{B}{x - 1} + \frac{C}{(x - 1)^{2}} \\ \Rightarrow 5 x^{2} - 8 x + 1 = A (x - 1)^{2} + B (2 x) (x - 1) + C (2 x) \\ x = 0 \Rightarrow A = 1 \\ x = 1 \Rightarrow C = - 1 \\ x = - 1 \Rightarrow 14 = 4 A + 4 B - 2 C \Rightarrow B = 2 \\ \int_{4}^{9} \frac{5 x^{2} - 8 x + 1}{2 x (x - 1)^{2}} d x = \int_{4}^{9} (\frac{1}{2 x} + \frac{2}{x - 1} + \frac{- 1}{(x - 1)^{2}}) d x \\ = {(\frac{1}{2} \ln | x | + 2 \ln | x - 1 | + \frac{1}{x - 1}) |}_{4}^{9} \\ = \frac{1}{2} \ln 9 + 2 \ln 8 + \frac{1}{8} - \frac{1}{2} \ln 4 - 2 \ln 3 - \frac{1}{3} \\ = \ln 3 + \ln 64 + \frac{1}{8} - \ln 2 - \ln 9 - \frac{1}{3} \\ = \ln \frac{3 (64)}{2 (9)} - \frac{5}{24} = \ln \frac{32}{3} - \frac{5}{24} \end{aligned}$

(36) تبرير: أثبت أن: $\int_{9}^{16} \frac{2 \sqrt{x}}{x - 4} d x = 4 (1 + \ln (\frac{5}{3}))$ .

$\begin{aligned} u = \sqrt{x} \Rightarrow u^{2} = x \Rightarrow d x = 2 u d u \\ x = 9 \Rightarrow u = 3 \\ x = 16 \Rightarrow u = 4 \\ \int_{9}^{16} \frac{2 \sqrt{x}}{x - 4} d x = \int_{3}^{4} \frac{2 u}{u^{2} - 4} 2 u d u = \int_{3}^{4} \frac{4 u^{2}}{u^{2} - 4} d u \\ = \int_{3}^{4} (4 + \frac{16}{u^{2} - 4}) d u \\ \frac{16}{u^{2} - 4} = \frac{16}{(u - 2) (u + 2)} = \frac{A}{u - 2} + \frac{B}{u + 2} \\ \Rightarrow 16 = A (u + 2) + B (u - 2) \\ u = 2 \Rightarrow A = 4 \\ u = - 2 \Rightarrow B = - 4 \\ \int_{3}^{4} (4 + \frac{16}{u^{2} - 4}) d u = \int_{3}^{4} (4 + \frac{4}{u - 2} + \frac{- 4}{u + 2}) d u \\ = {(4 u + 4 \ln | u - 2 | - 4 \ln | u + 2 |) |}_{3}^{4} \\ = 16 + 4 \ln 2 - 4 \ln 6 - 12 - 4 \ln 1 + 4 \ln 5 \\ = 4 + 4 \ln \frac{5}{3} = 4 (1 + \ln \frac{5}{3}) \\ \Rightarrow \int_{9}^{16} \frac{2 \sqrt{x}}{x - 4} d x = 4 (1 + \ln \frac{5}{3}) \end{aligned}$

(37) تبرير: أثبت أن: $\int_{0}^{1} \frac{4 x^{2} + 9 x + 4}{2 x^{2} + 5 x + 3} d x = 2 + \frac{1}{2} \ln \frac{5}{12}$ .

$\begin{aligned} \frac{4 x^{2} + 9 x + 4}{2 x^{2} + 5 x + 3} = 2 - \frac{x + 2}{2 x^{2} + 5 x + 3} \\ \frac{x + 2}{2 x^{2} + 5 x + 3} = \frac{x + 2}{(x + 1) (2 x + 3)} = \frac{A}{x + 1} + \frac{B}{2 x + 3} \\ \Rightarrow x + 2 = A (2 x + 3) + B (x + 1) \\ x = - 1 \Rightarrow A = 1 \\ x = - \frac{3}{2} \Rightarrow B = - 1 \\ \int_{0}^{1} \frac{4 x^{2} + 9 x + 4}{2 x^{2} + 5 x + 3} d x = \int_{0}^{1} (2 - \frac{1}{x + 1} + \frac{1}{2 x + 3}) d x \\ = {(2 x - \ln | x + 1 | + \frac{1}{2} \ln | 2 x + 3 |) |}_{0}^{1} \\ = 2 - \ln 2 + \frac{1}{2} \ln 5 - 0 + \ln 1 - \frac{1}{2} \ln 3 \\ = 2 - \frac{1}{2} \ln 4 + \frac{1}{2} \ln 5 - \frac{1}{2} \ln 3 \\ = 2 + \frac{1}{2} (\ln 5 - \ln 4 - \ln 3) = 2 + \frac{1}{2} \ln \frac{5}{12} \end{aligned}$

تحد: أجد كلاً من التكاملات الآتية:

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

$\begin{aligned} \int \frac{\sqrt{1 + \sqrt{x}}}{x} d x \\ u = \sqrt{1 + \sqrt{x}} \Rightarrow \frac{d u}{d x} = \frac{1}{4 \sqrt{x} \sqrt{1 + \sqrt{x}}}, 1 + \sqrt{x} = \\ \Rightarrow d x = 4 \sqrt{x} \sqrt{1 + \sqrt{x}} d u = 4 u (u^{2} - 1) d u \\ \int \frac{\sqrt{1 + \sqrt{x}}}{x} d x = \int \frac{u}{{(u^{2} - 1)}^{2}} 4 u (u^{2} - 1) d u = \end{aligned} \begin{aligned} \frac{4 u^{2}}{u^{2} - 1} = 4 + \frac{4}{u^{2} - 1} \\ \frac{4}{u^{2} - 1} = \frac{4}{(u - 1) (u + 1)} = \frac{A}{u - 1} + \frac{B}{u + 1} \\ \Rightarrow 4 = A (u + 1) + B (u - 1) \\ u = 1 \Rightarrow A = 2 \\ u = - 1 \Rightarrow B = - 2 \end{aligned} \begin{aligned} \begin{aligned} \int \frac{4 u^{2}}{u^{2} - 1} d u & = \int (4 + \frac{2}{u - 1} + \frac{- 2}{u + 1}) d u \\ = 4 u + 2 l n | u - 1 | - 2 l n | u + 1 | + C \\ = 4 u + 2 l n | \frac{u - 1}{u + 1} | + C \end{aligned} \\ \Rightarrow \int \frac{\sqrt{1 + \sqrt{x}}}{x} d x = 4 \sqrt{1 + \sqrt{x}} + 2 l n | \frac{\sqrt{1 + \sqrt{x}} - 1}{\sqrt{1 + \sqrt{x}} + 1} | + C \end{aligned}$

$\int \frac{x}{16 x^{4} - 1} d x$ (39)

$\begin{aligned} \frac{x}{16 x^{4} - 1} = \frac{x}{(4 x^{2} + 1) (2 x - 1) (2 x + 1)} = \frac{A x + B}{4 x^{2} + 1} + \frac{C}{2 x - 1} + \frac{D}{2 x + 1} \\ \Rightarrow x = (A x + B) (2 x - 1) (2 x + 1) + C (4 x^{2} + 1) (2 x + 1) + D (4 x^{2} + 1) (2 x - 1) \\ x = \frac{1}{2} \Rightarrow C = \frac{1}{8} \\ x = - \frac{1}{2} \Rightarrow D = \frac{1}{8} \\ x = 0 \Rightarrow 0 = - B + C - D \Rightarrow B = 0 \\ x = 1 \Rightarrow 1 = 3 A + 3 B + 15 C + 5 D \Rightarrow A = - \frac{1}{2} \\ \int \frac{x}{16 x^{4} - 1} d x = \int (\frac{- \frac{1}{2} x}{4 x^{2} + 1} + \frac{\frac{1}{8}}{2 x - 1} + \frac{\frac{1}{8}}{2 x + 1}) d x \\ = - \frac{1}{16} l n (4 x^{2} + 1) + \frac{1}{16} l n | 2 x - 1 | + \frac{1}{16} l n | 2 x + 1 | + C \\ = \frac{1}{16} l n | \frac{4 x^{2} - 1}{4 x^{2} + 1} | + C \end{aligned}$

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

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

إعداد : شبكة منهاجي التعليمية

12 / 02 / 2023

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