أتدرب وأحل المسائل

أتدرب وأحل المسائل

التكامل بالكسور الجزئية

أجد كلاً من التكاملات الآتية:

$\int \frac{x - 10}{x (x + 5)} d x$ (1)

$\begin{aligned} \frac{x - 10}{x (x + 5)} = \frac{A}{x} + \frac{B}{x + 5} \\ \Rightarrow x - 10 = A (x + 5) + B x \\ x = 0 \Rightarrow A = - 2 \\ \begin{aligned} x = - 5 \Rightarrow B = 3 \end{aligned} \\ \int \frac{x - 10}{x (x + 5)} d x = \int (\frac{- 2}{x} + \frac{3}{x + 5}) d x \\ = - 2 \ln | x | + 3 \ln | x + 5 | + C \end{aligned}$

$\int \frac{2}{1 - x^{2}} d x$ (2)

$\begin{aligned} \frac{2}{1 - x^{2}} = \frac{2}{(1 - x) (1 + x)} = \frac{A}{1 - x} + \frac{B}{1 + x} \\ \Rightarrow 2 = A (1 + x) + B (1 - x) \\ x = 1 \Rightarrow A = 1 \\ \begin{aligned} x = - 1 \Rightarrow B & = 1 \\ \int \frac{2}{1 - x^{2}} d x & = \int (\frac{1}{1 - x} + \frac{1}{1 + x}) d x \\ = - \ln | 1 - x | + \ln | 1 + x | + C \\ = \ln | \frac{1 + x}{1 - x} | + C \end{aligned} \end{aligned}$

$\int \frac{4}{(x - 2) (x - 4)} d x$ (3)

$\begin{aligned} \frac{4}{(x - 2) (x - 4)} = \frac{A}{x - 2} + \frac{B}{x - 4} \\ \Rightarrow 4 = A (x - 4) + B (x - 2) \\ x = 2 \Rightarrow A = - 2 \\ x = 4 \Rightarrow B = 2 \\ \int \frac{4}{(x - 2) (x - 4)} d x = \int (\frac{- 2}{x - 2} + \frac{2}{x - 4}) d x \\ = - 2 \ln | x - 2 | + 2 \ln | x - 4 | + C \\ = 2 \ln | \frac{x - 4}{x - 2} | + C \end{aligned}$

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

$\begin{aligned} \frac{3 x + 4}{x^{2} + x} = \frac{3 x + 4}{x (x + 1)} = \frac{A}{x} + \frac{B}{x + 1} \\ \Rightarrow 3 x + 4 = A (x + 1) + B x \\ x = 0 \Rightarrow A = 4 \\ x = - 1 \Rightarrow B = - 1 \\ \begin{aligned} \frac{3 x + 4}{x^{2} + x} d x & = \int (\frac{4}{x} + \frac{- 1}{x + 1}) d x \\ = 4 \ln | x | - \ln | x + 1 | + C \end{aligned} \end{aligned}$

$\int \frac{x^{2}}{x^{2} - 4} d x$ (5)

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

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

$\begin{aligned} \frac{3 x - 6}{x^{2} + x - 2} = \frac{3 x - 6}{(x + 2) (x - 1)} = \frac{A}{x + 2} + \frac{B}{x - 1} \\ \Rightarrow 3 x - 6 = A (x - 1) + B (x + 2) \\ x = - 2 \Rightarrow A = 4 \\ x = 1 \Rightarrow B = - 1 \\ \begin{aligned} x \frac{3 x - 6}{x^{2} + x - 2} d x = \int (\frac{4}{x + 2} + \frac{- 1}{x - 1}) d x \\ = 4 \ln | x + 2 | - \ln | x - 1 | + C \end{aligned} \end{aligned}$

$\int \frac{4 x + 10}{4 x^{2} - 4 x - 3} d x$ (7)

$\begin{aligned} \frac{4 x + 10}{4 x^{2} - 4 x - 3} = \frac{4 x + 10}{(2 x - 3) (2 x + 1)} = \frac{A}{2 x - 3} + \frac{B}{2 x + 1} \\ \Rightarrow 4 x + 10 = A (2 x + 1) + B (2 x - 3) \\ x = \frac{3}{2} \Rightarrow A = 4 \\ x = - \frac{1}{2} \Rightarrow B = - 2 \\ \int \frac{4 x + 10}{4 x^{2} - 4 x - 3} d x = \int (\frac{4}{2 x - 3} + \frac{- 2}{2 x + 1}) d x \\ = 2 \ln | 2 x - 3 | - \ln | 2 x + 1 | + C \end{aligned}$

$\int \frac{2 x^{2} + 9 x - 11}{x^{3} + 2 x^{2} - 5 x - 6} d x$ (8)

$\begin{aligned} \frac{2 x^{2} + 9 x - 11}{x^{3} + 2 x^{2} - 5 x - 6} = \frac{2 x^{2} + 9 x - 11}{(x - 2) (x + 1) (x + 3)} = \frac{A}{x - 2} + \frac{B}{x + 1} + \frac{C}{x + 3} \\ \Rightarrow 2 x^{2} + 9 x - 11 = A (x + 1) (x + 3) + B (x - 2) (x + 3) + C (x - 2) (x + 1) \\ x = 2 \Rightarrow A = 1 \\ x = - 1 \Rightarrow B = 3 \\ x = - 3 \Rightarrow C = - 2 \\ \int \frac{2 x^{2} + 9 x - 11}{x^{3} + 2 x^{2} - 5 x - 6} d x = \int (\frac{1}{x - 2} + \frac{3}{x + 1} + \frac{- 2}{x + 3}) d x \\ = \ln | x - 2 | + 3 \ln | x + 1 | - 2 \ln | x + 3 | + C \end{aligned}$

$\int \frac{4 x}{x^{2} - 2 x - 3} d x$ (9)

$\begin{aligned} \frac{4 x}{x^{2} - 2 x - 3} = \frac{4 x}{(x - 3) (x + 1)} = \frac{A}{x - 3} + \frac{B}{x + 1} \\ \Rightarrow 4 x = A (x + 1) + B (x - 3) \\ x = 3 \Rightarrow A = 3 \\ x = - 1 \Rightarrow B = 1 \\ \int \frac{4 x}{x^{2} - 2 x - 3} d x = \int (\frac{3}{x - 3} + \frac{1}{x + 1}) d x \\ = 3 \ln | x - 3 | + \ln | x + 1 | + C \end{aligned}$

$\int \frac{8 x^{2} - 19 x + 1}{(2 x + 1) (x - 2)^{2}} d x$ (10)

$\begin{aligned} \frac{8 x^{2} - 19 x + 1}{(2 x + 1) (x - 2)^{2}} = \frac{A}{2 x + 1} + \frac{B}{x - 2} + \frac{C}{(x - 2)^{2}} \\ \Rightarrow 8 x^{2} - 19 x + 1 = A (x - 2)^{2} + B (2 x + 1) (x - 2) + C (2 x + 1) \\ x = - \frac{1}{2} \Rightarrow A = 2 \\ x = 2 \Rightarrow C = - 1 \\ x = 0 \Rightarrow 1 = 4 A - 2 B + C \Rightarrow B = 3 \\ \int \frac{8 x^{2} - 19 x + 1}{(2 x + 1) (x - 2)^{2}} d x = \int (\frac{2}{2 x + 1} + \frac{3}{x - 2} + \frac{- 1}{(x - 2)^{2}}) d x \\ = \ln | 2 x + 1 | + 3 \ln | x - 2 | + \frac{1}{x - 2} + C \end{aligned}$

$\int \frac{9 x^{2} - 3 x + 2}{9 x^{2} - 4} d x$ (11)

$\begin{aligned} \int \frac{9 x^{2} - 3 x + 2}{9 x^{2} - 4} d x = \int (1 + \frac{6 - 3 x}{9 x^{2} - 4}) d x \\ \frac{6 - 3 x}{9 x^{2} - 4} = \frac{6 - 3 x}{(3 x - 2) (3 x + 2)} = \frac{A}{3 x - 2} + \frac{B}{3 x + 2} \\ \Rightarrow 6 - 3 x = A (3 x + 2) + B (3 x - 2) \\ x = \frac{2}{3} \Rightarrow A = 1 \\ x = - \frac{2}{3} \Rightarrow B = - 2 \\ \int \frac{9 x^{2} - 3 x + 2}{9 x^{2} - 4} d x = \int (1 + \frac{1}{3 x - 2} + \frac{- 2}{3 x + 2}) d x \\ = x + \frac{1}{3} \ln | 3 x - 2 | - \frac{2}{3} \ln | 3 x + 2 | + C \end{aligned}$

$\int \frac{x^{3} + 2 x^{2} + 2}{x^{2} + x} d x$ (12)

$\begin{aligned} \int \frac{x^{3} + 2 x^{2} + 2}{x^{2} + x} d x = \int (x + 1 + \frac{2 - x}{x^{2} + x}) d x \\ \frac{2 - x}{x^{2} + x} = \frac{2 - x}{x (x + 1)} = \frac{A}{x} + \frac{B}{x + 1} \\ \Rightarrow 2 - x = A (x + 1) + B x \\ x = 0 \Rightarrow A = 2 \\ x = - 1 \Rightarrow B = - 3 \\ \int \frac{x^{3} + 2 x^{2} + 2}{x^{2} + x} d x = \int (x + 1 + \frac{2}{x} + \frac{- 3}{x + 1}) d x \\ = \frac{1}{2} x^{2} + x + 2 \ln | x | - 3 \ln | x + 1 | + C \end{aligned}$

$\int \frac{x^{2} + x + 2}{3 - 2 x - x^{2}} d x$ (13)

$\begin{aligned} \int \frac{x^{2} + x + 2}{3 - 2 x - x^{2}} d x = \int (- 1 + \frac{5 - x}{- x^{2} - 2 x + 3}) d x \\ \frac{5 - x}{- x^{2} - 2 x + 3} = \frac{x - 5}{(x + 3) (x - 1)} = \frac{A}{x + 3} + \frac{B}{x - 1} \\ \Rightarrow x - 5 = A (x - 1) + B (x + 3) \\ x = - 3 \Rightarrow A = 2 \\ \begin{aligned} x = 1 \Rightarrow B = - 1 \end{aligned} \\ \int \frac{x^{2} + x + 2}{3 - 2 x - x^{2}} d x = \int (- 1 + \frac{2}{x + 3} + \frac{- 1}{x - 1}) d x \\ = - x + 2 \ln | x + 3 | - \ln | x - 1 | + C \end{aligned}$

$\int \frac{2 x - 4}{(x^{2} + 4) (x + 2)} d x$ (14)

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

$\int \frac{x^{3} - 4 x^{2} - 2}{x^{3} + x^{2}} d x$ (15)

$\begin{aligned} \int \frac{x^{3} - 4 x^{2} - 2}{x^{3} + x^{2}} d x = \int (1 + \frac{- 5 x^{2} - 2}{x^{3} + x^{2}}) d x \\ \frac{- 5 x^{2} - 2}{x^{3} + x^{2}} = \frac{- 5 x^{2} - 2}{x^{2} (x + 1)} = \frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{x + 1} \\ \Rightarrow - 5 x^{2} - 2 = A x (x + 1) + B (x + 1) + C x^{2} \\ x = 0 \Rightarrow B = - 2 \\ x = - 1 \Rightarrow C = - 7 \\ x = 1 \Rightarrow - 7 = 2 A + 2 B + C \Rightarrow A = 2 \\ \int \frac{x^{3} - 4 x^{2} - 2}{x^{3} + x^{2}} d x = \int (1 + \frac{2}{x} + \frac{- 2}{x^{2}} + \frac{- 7}{x + 1}) d x \\ = x + 2 \ln | x | + \frac{2}{x} - 7 \ln | x + 1 | + C \end{aligned}$

$\int \frac{3 - x}{2 - 5 x - 12 x^{2}} d x$ (16)

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

$\int \frac{3 x^{3} - x^{2} + 12 x - 6}{x^{4} + 6 x^{2}} d x$ (17)

$\begin{aligned} \frac{3 x^{3} - x^{2} + 12 x - 6}{x^{4} + 6 x^{2}} = \frac{3 x^{3} - x^{2} + 12 x - 6}{x^{2} (x^{2} + 6)} = \frac{A}{x} + \frac{B}{x^{2}} + \frac{C x + D}{x^{2} + 6} \\ \Rightarrow 3 x^{3} - x^{2} + 12 x - 6 = A x (x^{2} + 6) + B (x^{2} + 6) + (C x + D) (x^{2}) \\ x = 0 \Rightarrow B = - 1 \\ x = 1 \Rightarrow 8 = 7 A + 7 B + C + D \dots \dots \dots \dots \dots (1) \\ x = - 1 \Rightarrow - 22 = - 7 A + 7 B - C + D \dots \dots (2) \\ x = 2 \Rightarrow 38 = 20 A + 10 B + 8 C + 4 D \dots \dots (3) \end{aligned}$

بجمع (1)، (2) ينتج أن: $14 B + 2 D = - 14$ ، وبتعويض $B = - 1$ ، نجد أن $D = 0$

وبطرح (2) من (1) ينتج أن $14 A + 2 C = 30$ أي أن $C = 15 - 7 A$

بالتعويض في (3) ينتج أن:

$\begin{aligned} 20 A - 10 + 8 (15 - 7 A) & = 38 \\ - 36 A & = - 72 \Rightarrow A = 2 \\ C & = 15 - 7 (2) = 1 \end{aligned} \begin{array}{r} \int \frac{3 x^{3} - x^{2} + 12 x - 6}{x^{4} + 6 x^{2}} d x = \int (\frac{2}{x} + \frac{- 1}{x^{2}} + \frac{x}{x^{2} + 6}) d x \\ = 2 l n | x | + \frac{1}{x} + \frac{1}{2} l n | x^{2} + 6 | + C \end{array}$

$\int \frac{5 x - 2}{(x - 2)^{2}} d x$ (18)

$\begin{aligned} \frac{5 x - 2}{(x - 2)^{2}} = \frac{A}{x - 2} + \frac{B}{(x - 2)^{2}} \\ \Rightarrow 5 x - 2 = A (x - 2) + B \\ x = 2 \Rightarrow B = 8 \\ x = 0 \Rightarrow - 2 = - 2 A + B \Rightarrow A = 5 \\ \int \frac{5 x - 2}{(x - 2)^{2}} d x = \int (\frac{5}{x - 2} + \frac{8}{(x - 2)^{2}}) d x \\ = 5 \ln | x - 2 | - \frac{8}{x - 2} + C \end{aligned}$

ملاحظة: يمكن حل هذا التكامل بالتعويض $u = x - 2$

كما يمكن حله بالأجزاء حيث: $u = 5 x - 2, d v = (x - 2)^{- 2}$

أجد قيمة كل من التكاملات الآتية:

$\int_{2}^{4} \frac{6 + 3 x - x^{2}}{x^{3} + 2 x^{2}} d x$ (19)

$\begin{aligned} \frac{6 + 3 x - x^{2}}{x^{3} + 2 x^{2}} = \frac{6 + 3 x - x^{2}}{x^{2} (x + 2)} = \frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{x + 2} \\ \Rightarrow 6 + 3 x - x^{2} = A x (x + 2) + B (x + 2) + C (x^{2}) \\ x = 0 \Rightarrow B = 3 \\ x = - 2 \Rightarrow C = - 1 \\ x = 1 \Rightarrow 8 = 3 A + 3 B + C \Rightarrow A = 0 \\ \begin{aligned} \int_{2}^{4} \frac{6 + 3 x - x^{2}}{x^{3} + 2 x^{2}} d x & = \int_{2}^{4} (\frac{3}{x^{2}} + \frac{- 1}{x + 2}) d x \\ = {(- \frac{3}{x} - \ln | x + 2 |) |}_{2}^{4} \\ = - \frac{3}{4} - \ln 6 + \frac{3}{2} + \ln 4 = \frac{3}{4} + \ln \frac{2}{3} \end{aligned} \end{aligned}$

$\int_{- 1 / 3}^{1 / 3} \frac{9 x^{2} + 4}{9 x^{2} - 4} d x$ (20)

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

$\int_{0}^{1} \frac{17 - 5 x}{(2 x + 3) (2 - x)^{2}} d x$ (21)

$\begin{aligned} \frac{17 - 5 x}{(2 x + 3) (2 - x)^{2}} = \frac{A}{2 x + 3} + \frac{B}{2 - x} + \frac{C}{(2 - x)^{2}} \\ \Rightarrow 17 - 5 x = A (2 - x)^{2} + B (2 - x) (2 x + 3) + C (2 x + 3) \\ x = - \frac{3}{2} \Rightarrow A = 2 \\ x = 2 \Rightarrow C = 1 \\ x = 0 \Rightarrow 17 = 4 A + 6 B + 3 C \Rightarrow B = 1 \\ \int_{0}^{1} \frac{17 - 5 x}{(2 x + 3) (2 - x)^{2}} d x = \int_{0}^{1} (\frac{2}{2 x + 3} + \frac{1}{2 - x} + \frac{1}{(2 - x)^{2}}) d x \\ = {(l n | 2 x + 3 | - l n | 2 - x | + \frac{1}{2 - x}) |}_{0}^{1} \\ = l n 5 + 1 - l n 3 + l n 2 - \frac{1}{2} = \frac{1}{2} + l n \frac{10}{3} \end{aligned}$

$\int_{1}^{4} \frac{4}{16 x^{2} + 8 x - 3} d x$ (22)

$\begin{aligned} \frac{4}{16 x^{2} + 8 x - 3} = \frac{4}{(4 x - 1) (4 x + 3)} = \frac{A}{4 x - 1} + \frac{B}{4 x + 3} \\ \Rightarrow 4 = A (4 x + 3) + B (4 x - 1) \\ x = \frac{1}{4} \Rightarrow A = 1 \\ x = - \frac{3}{4} \Rightarrow B = - 1 \\ \int_{1}^{4} \frac{4}{16 x^{2} + 8 x - 3} d x = \int_{1}^{4} (\frac{1}{4 x - 1} + \frac{- 1}{4 x + 3}) d x \\ = {(\frac{1}{4} \ln | 4 x - 1 | - \frac{1}{4} \ln | 4 x + 3 |) |}_{1}^{4} \\ = {(\frac{1}{4} \ln | \frac{4 x - 1}{4 x + 3} |) |}_{1}^{4} \\ = \frac{1}{4} (\ln \frac{15}{19} - \ln \frac{3}{7}) = \frac{1}{4} \ln \frac{35}{19} \end{aligned}$

$\int_{3}^{4} \frac{5 x + 5}{x^{2} + x - 6} d x$ (23)

$\begin{aligned} \frac{5 x + 5}{x^{2} + x - 6} = \frac{5 x + 5}{(x - 2) (x + 3)} = \frac{A}{x - 2} + \frac{B}{x + 3} \\ \Rightarrow 5 x + 5 = A (x + 3) + B (x - 2) \\ x = 2 \Rightarrow A = 3 \\ x = - 3 \Rightarrow B = 2 \\ \int_{3}^{4} \frac{5 x + 5}{x^{2} + x - 6} d x = \int_{3}^{4} (\frac{3}{x - 2} + \frac{2}{x + 3}) d x \\ = {(3 \ln | x - 2 | + 2 \ln | x + 3 |) |}_{3}^{4} \\ = 3 \ln 2 + 2 \ln 7 - 2 \ln 6 = \ln \frac{98}{9} \end{aligned}$

$\int_{3}^{4} \frac{4}{x^{3} - 4 x^{2} + 4 x} d x$ (24)

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

أجد مساحة المنطقة المظللة في كل من التمثيلين البيانيين الآتيين:

$\begin{aligned} A = \int_{0}^{1} \frac{1}{x^{2} - 5 x + 6} d x \\ \frac{1}{x^{2} - 5 x + 6} = \frac{1}{(x - 3) (x - 2)} = \frac{A}{x - 3} + \frac{B}{x - 2} \\ \Rightarrow 1 = A (x - 2) + B (x - 3) \\ x = 3 \Rightarrow A = 1 \\ x = 2 \Rightarrow B = - 1 \\ A = \int_{0}^{1} \frac{1}{x^{2} - 5 x + 6} d x = \int_{0}^{1} (\frac{1}{x - 3} + \frac{- 1}{x - 2}) d x \\ = {(\ln | x - 3 | - \ln | x - 2 |) |}_{0}^{1} \\ = {\ln | \frac{x - 3}{x - 2} | |}_{0}^{1} \\ = \ln 2 - \ln \frac{3}{2} = \ln \frac{4}{3} \end{aligned}$

$\begin{aligned} A = \int_{1}^{2} \frac{x^{2} + 1}{3 x - x^{2}} d x \\ \frac{x^{2} + 1}{3 x - x^{2}} = - 1 + \frac{3 x + 1}{3 x - x^{2}} \\ \frac{3 x + 1}{3 x - x^{2}} = \frac{3 x + 1}{x (3 - x)} = \frac{A}{x} + \frac{B}{3 - x} \\ \Rightarrow 3 x + 1 = A (3 - x) + B x \\ x = 0 \Rightarrow A = \frac{1}{3} = - 2 + \frac{1}{3} \ln 2 + 1 + \frac{10}{3} \ln 2 \\ x = 3 \Rightarrow B = - \frac{11}{3} \ln 2 \\ \int_{1}^{2} \frac{x^{2} + 1}{3 x - x^{2}} d x = \int_{1}^{2} (- 1 + \frac{10}{3} + \frac{1}{3 - x}) d x \end{aligned}$

يبين الشكل المجاور جزءاً من منحنى الاقتران: $f (x) = \frac{4 x - 5}{2 x^{2} - 5 x - 3}$ :

(27) أجد إحداثيي النقطة $A$ .

$f (x) = 0 \Rightarrow 4 x - 5 = 0 \Rightarrow x = \frac{5}{4} \Rightarrow A (\frac{5}{4}, 0)$

(28) أجد مساحة المنطقة المظللة.

$A = \int_{0}^{\frac{5}{4}} \frac{4 x - 5}{2 x^{2} - 5 x - 3} d x = \ln {| 2 x^{2} - 5 x - 3 |}_{0}^{\frac{5}{4}} = \ln \frac{49}{8} - \ln 3 = \ln \frac{49}{24}$

ملاحظة: البسط هو مشتقة المقام، فلا داعي لتجزئة الكسر.

أجد كلاً من التكاملات الآتية:

$\int \frac{\sin x}{\cos x + \cos^{2} x} d x$ (29)

$\begin{aligned} u = \cos x \Rightarrow \frac{d u}{d x} = - \sin x \Rightarrow d x = \frac{d u}{- \sin x} \\ \int \frac{\sin x}{\cos x + \cos^{2} x} d x = \int \frac{\sin x}{u + u^{2}} \times \frac{d u}{- \sin x} = \int \frac{- 1}{u + u^{2}} d u \\ \frac{- 1}{u + u^{2}} = \frac{- 1}{u (1 + u)} = \frac{A}{u} + \frac{B}{1 + u} \\ \Rightarrow - 1 = A (1 + u) + B u \\ u = 0 \Rightarrow A = - 1 \\ u = - 1 \Rightarrow B = 1 \\ \int \frac{- 1}{u + u^{2}} d u = \int (\frac{- 1}{u} + \frac{1}{1 + u}) d u \\ \Rightarrow \int \frac{\sin x}{\cos x + \cos^{2} x} d x = - \ln | \cos x | + \ln | 1 + \cos x | + C \\ = \ln | 1 + u | + C \\ \frac{1 + \cos x}{\cos x} | + C = \ln | 1 + \sec x ∣ + C \end{aligned}$

$\int \frac{1}{x^{2} + x \sqrt{x}} d x$ (30)

$\begin{aligned} u = \sqrt{x} \Rightarrow u^{2} = x \Rightarrow d x = 2 u d u \\ \int \frac{1}{x^{2} + x \sqrt{x}} d x = \int \frac{1}{u^{4} + u^{3}} 2 u d u = \int \frac{2}{u^{3} + u^{2}} d u \\ \frac{2}{u^{3} + u^{2}} = \frac{2}{u^{2} (u + 1)} = \frac{A}{u} + \frac{B}{u^{2}} + \frac{C}{u + 1} \\ \Rightarrow 2 = A u (u + 1) + B (u + 1) + C u^{2} \\ u = 0 \Rightarrow B = 2 \\ u = - 1 \Rightarrow C = 2 \\ u = 1 \Rightarrow 2 = 2 A + 2 B + C \Rightarrow A = - 2 \\ \int \frac{2}{u^{3} + u^{2}} d u = \int (\frac{- 2}{u} + \frac{2}{u^{2}} + \frac{2}{u + 1}) d u \\ = - 2 \ln | u | - \frac{2}{u} + 2 \ln | u + 1 | + C \\ \Rightarrow \int \frac{1}{x^{2} + x \sqrt{x}} d x = 2 \ln | \frac{u + 1}{u} | - \frac{2}{u} + C \end{aligned}$

$\int \frac{e^{2 x}}{e^{2 x} + 3 e^{x} + 2} d x$ (31)

$\begin{aligned} u = e^{x} \Rightarrow \frac{d u}{d x} = e^{x} = u ⟹ d x = \frac{d u}{u} \\ \int \frac{e^{2 x}}{e^{2 x} + 3 e^{x} + 2} d x = \int \frac{u^{2}}{u^{2} + 3 u + 2} \times \frac{d u}{u} = \int \frac{u}{u^{2} + 3 u + 2} d u \\ \frac{u}{u^{2} + 3 u + 2} = \frac{u}{(u + 1) (u + 2)} = \frac{A}{u + 1} + \frac{B}{u + 2} \\ \Rightarrow u = A (u + 2) + B (u + 1) \\ u = - 1 \Rightarrow A = - 1 \\ u = - 2 \Rightarrow B = 2 \\ \int \frac{u}{u^{2} + 3 u + 2} d u = \int (\frac{- 1}{u + 1} + \frac{2}{u + 2}) d u \\ = - \ln | u + 1 | + 2 \ln | u + 2 | + C \\ \Rightarrow \int \frac{e^{2 x}}{e^{2 x} + 3 e^{x} + 2} d x = - \ln (e^{x} + 1) + 2 \ln (e^{x} + 2) + C \end{aligned}$

$\int \frac{\cos x}{\sin x (\sin^{2} x - 4)} d x$ (32)

$\begin{aligned} u = \sin x \Rightarrow \frac{d u}{d x} = \cos x \Rightarrow d x = \frac{d u}{\cos x} \\ \int \frac{\cos x}{\sin x (\sin^{2} x - 4)} d x = \int \frac{\cos x}{u (u^{2} - 4)} \times \frac{d u}{\cos x} = \int \frac{1}{u (u^{2} - 4)} d u \\ \frac{1}{u (u^{2} - 4)} = \frac{1}{u (u - 2) (u + 2)} = \frac{A}{u} + \frac{B}{u - 2} + \frac{C}{u + 2} \\ \Rightarrow 1 = A (u - 2) (u + 2) + B u (u + 2) + C u (u - 2) \\ u = 0 \Rightarrow A = - \frac{1}{4} \\ u = 2 \Rightarrow B = \frac{1}{8} \\ u = - 2 \Rightarrow C = \frac{1}{8} \\ \int \frac{1}{u (u^{2} - 4)} d u = \int (\frac{- \frac{1}{4}}{u} + \frac{\frac{1}{8}}{u - 2} + \frac{\frac{1}{8}}{u + 2}) d u \\ = - \frac{1}{4} \ln | u | + \frac{1}{8} \ln | u - 2 | + \frac{1}{8} \ln | u + 2 | + C \\ \Rightarrow \int \frac{\cos x}{\sin x (\sin^{2} x - 4)} d x \\ = - \frac{1}{4} \ln | \sin x | + \frac{1}{8} \ln | \sin x - 2 | + \frac{1}{8} \ln | \sin x + 2 | + C \end{aligned}$

إعداد : شبكة منهاجي التعليمية

12 / 02 / 2023

حمّل تطبيق منهاجي الجديد

النقاشات

حمّل تطبيق منهاجي الجديد

النقاشات

جميع الحقوق محفوظة © 2025