رياضيات فصل ثاني

التوجيهي علمي

الشرح الملخص أوراق العمل حل اسئلة الدرس النتاجات

الدرس الرابع: التكامل بالأجزاء .

التحقق من الفهم ص 63

أجد كلاً من التكاملات الآتية:

$a \int x \sin x d x S o l u t i o n 1 : B y l a w u = x d v = \sin x d u = d x v = \int \sin x d x = - \cos x \int x s i n x d x = - x c o s x + \int c o s x d x = - x c o s x + s i n x + c S o l u t i o n 2 : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ s i n x & x & + \\ - c o s x & 1 & - \\ - s i n x & 0 & + \end{matrix} \int x s i n x d x = - x c o s x + \sin x + c$

$b \int x^{2} \ln x d x S o l u t i o n 1 : B y l a w u = \ln x d v = x^{2} d u = \frac{1}{x} v = \frac{x^{3}}{3} \int x^{2} l n x d x = \frac{x^{3}}{3} l n x - \int \frac{x^{3}}{3} \times \frac{1}{x} d x = \frac{x^{3}}{3} l n x - \frac{1}{3} \int x^{2} d x = \frac{x^{3}}{3} l n x + \frac{x^{3}}{9} + c S o l u t i o n 2 : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ x^{2} & \ln x & + \\ \frac{x^{3}}{3} & \frac{1}{x} & - \end{matrix} \int x^{2} l n x d x = \frac{x^{3}}{3} \ln x - \frac{1}{3} \int x^{2} d x = \frac{x^{3}}{3} l n x - \frac{x^{3}}{9} + c$

$c \int 2 x \sqrt{7 - 3 x} d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ {(7 - 3 x)}^{\frac{1}{2}} & 2 x & + \\ - \frac{2}{9} {(7 - 3 x)}^{\frac{3}{2}} & 2 & - \\ \frac{4}{135} {(7 - 3 x)}^{\frac{5}{2}} & 0 & + \end{matrix} \int 2 x \sqrt{7 - 3 x} d x = - \frac{4 x}{9} {(7 - 3 x)}^{\frac{3}{2}} - \frac{8}{135} {(7 - 3 x)}^{\frac{5}{2}} + c$

$d \int 3 x e^{4 x} d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ e^{4 x} & 3 x & + \\ \frac{e^{4 x}}{4} & 3 & - \\ \frac{e^{4 x}}{16} & 0 & + \end{matrix} \int 3 x e^{4 x} d x = \frac{3 x e^{4 x}}{4} - \frac{3 e^{4 x}}{16} + c$

التحقق من الفهم ص 64

أجد كلاً من التكاملين الآتيين:

$a \int x^{2} s i n x d x S o l u t i o n : B y T a b l e \begin{matrix} I n t & D i r & s i g n e \\ s i n x & x^{2} & + \\ - c o s x & 2 x & - \\ - s i n x & 2 & + \\ c o s x & 0 & - \end{matrix} \int x s i n x d x = - x^{2} c o s x + 2 x s i n x + 2 \cos x + c$

$b \int x^{3} e^{4 x} d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ e^{4 x} & x^{3} & + \\ \frac{e^{4 x}}{4} & 3 x^{2} & - \\ \frac{e^{4 x}}{16} & 6 x & + \\ \frac{e^{4 x}}{64} & 6 & - \\ \frac{e^{4 x}}{256} & 0 & + \end{matrix} \int x^{3} e^{4 x} d x = \frac{x^{3} e^{4 x}}{4} - \frac{3 x^{2} e^{4 x}}{16} + \frac{6 x e^{4 x}}{64} + \frac{6 e^{4 x}}{256} + c = \frac{x^{3} e^{4 x}}{4} - \frac{3 x^{2} e^{4 x}}{16} + \frac{3 x e^{4 x}}{32} - \frac{3 e^{4 x}}{128} + c$

التحقق من الفهم ص 66

أجد كلاً من التكاملين الآتيين:

$a \int \frac{\sin x}{e^{x}} d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ e^{- x} & \sin x & + \\ - e^{- x} & \cos x & - \\ e^{- x} & - \sin x & + \end{matrix} \int e^{- x} \sin x d x = - e^{- x} \sin x - e^{- x} \cos x - \int e^{- x} s i n x d x 2 \int e^{- x} s i n x d x = - e^{- x} s i n x - e^{- x} c o s x \int \frac{\sin x}{e^{x}} d x = \frac{- e^{- x} \sin x - e^{- x} \cos x}{2} + c$

$b \int s e c^{3} x d x S o l u t i o n : B y T a b u l a r m e t h o d \int s e c^{3} x d x = \int s e c^{2} x s e c x d x \begin{matrix} d v & u & s i g n e \\ s e c^{2} x & s e c x & + \\ \tan x & s e c x \tan x & - \end{matrix} \int s e c^{3} x d x = s e c x \tan x - \int \tan^{2} x s e c x d x = s e c x t a n x - \int (s e c^{2} x - 1) s e c x d x = s e c x t a n x - \int s e c^{3} x d x + \int s e c x d x b u t \int s e c x d x = \int s e c x \times \frac{s e c x + \tan x}{s e c x + t a n x} d x = \int \frac{s e c^{2} x + s e c x t a n x}{s e c x + t a n x} d x = \ln (s e c x + t a n x) 2 \int s e c^{3} x d x = s e c x t a n x + l n (s e c x + t a n x) \int s e c^{3} x d x = \frac{s e c x \tan x + \ln (s e c x + \tan x)}{2} + c$

التحقق من الفهم ص 67

أجد كلاً من التكاملين الآتيين:

$a \int x^{4} \cos x d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ \cos x & x^{4} & + \\ \sin x & 4 x^{3} & - \\ - \cos x & 12 x^{2} & + \\ - \sin x & 24 x & - \\ \cos x & 24 & + \\ \sin x & 0 & - \end{matrix} \int x^{4} c o s x d x = x^{4} s i n x + 4 x^{3} c o s x - 12 x^{2} s i n x + 24 x c o s x + 24 s i n x + c$

$b \int x^{5} e^{x} d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ e^{x} & x^{5} & + \\ e^{x} & 5 x^{4} & - \\ e^{x} & 20 x^{3} & + \\ e^{x} & 60 x^{2} & - \\ e^{x} & 120 x & + \\ e^{x} & 120 & - \\ e^{x} & 0 & + \end{matrix} \int x^{5} e^{x} d x = x^{5} e^{x} - 5 x^{4} e^{x} + 20 x^{3} e^{x} - 60 x^{2} e^{x} + 120 x e^{x} - 120 e^{x} + c$

التحقق من الفهم ص 69

$C' (x) = (0.1 x + 1) e^{0.03 x}, C (10) = 200 S o l u t i o n : C (x) = \int C' (x) d x B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ e^{0.03 x} & 0.1 x + 1 & + \\ \frac{e^{0.03 x}}{0.03} & 0.1 & - \\ \frac{e^{0.03 x}}{0.0009} & 0 & + \end{matrix} C (x) = (0.1 x + 1) \frac{e^{0.03 x}}{0.03} + 0.1 \times \frac{e^{0.03 x}}{0.0009} + c C (10) = (0.1 (10) + 1) \frac{e^{0.03 (10)}}{0.03} + \frac{0.1 e^{0.03 (10)}}{0.0009} + c = 200 2 \frac{e^{0.3}}{0.03} + 0.1 \times \frac{e^{0.3}}{0.0009} + c = 200 c = 200 - \frac{1.6 e^{0.3}}{0.009} C (x) = (0.1 x + 1) \frac{e^{0.03 x}}{0.03} + \frac{e^{0.03 x}}{0.009} + 200 - \frac{16 e^{0.3}}{0.09}$

التحقق من الفهم صفحة 70

$a \int_{1}^{e} \frac{l n x}{x^{2}} d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ x^{- 2} & l n x & + \\ \frac{- 1}{x} & \frac{1}{x} & - \end{matrix} \int \frac{l n x}{x^{2}} d x = - \frac{1}{x} l n x + \int x^{- 2} d x = - \frac{1}{x} l n x + \frac{- 1}{x} e^{1} = - \frac{l n x + 1}{x} e^{1} = - \frac{(l n e + 1)}{e} + (l n 1 + 1) = 1 - \frac{2}{e}$

$b \int_{0}^{1} x e^{- 2 x} d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ e^{- 2 x} & x & + \\ \frac{e^{- 2 x}}{- 2} & 1 & - \\ \frac{e^{- 2 x}}{4} & 0 & + \end{matrix} \int_{0}^{1} x e^{- 2 x} d x = - \frac{x e^{- 2 x}}{2} - \frac{e^{- 2 x}}{4} 1^{0} = - \frac{e^{- 2 x}}{2} (x + \frac{1}{2}) 1^{0} = - \frac{e^{- 2}}{2} (\frac{3}{2}) + \frac{1}{2} (\frac{1}{2}) = \frac{1 - 3 e^{- 2}}{4}$

التحقق من الفهم صفحة 70

$a \int (x^{3} + x^{5}) \sin (x^{2}) d x S o l u t i o n : l e t u = x^{2} \Rightarrow d x = \frac{d u}{2 x} \int (x^{3} + x^{5}) s i n (x^{2}) d x = \int x^{3} (1 + x^{2}) s i n u \frac{d u}{2 x} = \frac{1}{2} \int x^{2} (1 + x^{2}) s i n u d u = \frac{1}{2} \int u (1 + u) s i n u d u = \frac{1}{2} \int (u + u^{2}) s i n u d u B y T a b u l a r m e t h o d \begin{matrix} d v & w & s i g n e \\ s i n u & u + u^{2} & + \\ - c o s u & 1 + 2 u & - \\ - s i n u & 2 & + \end{matrix} \frac{1}{2} \int (u + u^{2}) s i n u d u = - \frac{1}{2} (u + u^{2}) (c o s u) + \frac{1}{2} (1 + 2 u) (s i n u) - \int s i n u d u = - \frac{1}{2} (u + u^{2}) (c o s u) + \frac{1}{2} (1 + 2 u) (\sin u) + \cos u + c b u t u = x^{2} : \int (x^{3} + x^{5}) s i n (x^{2}) d x = - \frac{1}{2} (x^{2} + x^{4}) (c o s x^{2}) + \frac{1}{2} (1 + 2 x^{2}) (s i n x^{2}) + c o s x^{2} + c$

$b \int x^{5} e^{x^{2}} d x S o l u t i o n : l e t u = x^{2} \Rightarrow d x = \frac{d u}{2 x} a n d x^{4} = u^{2} \int x^{5} e^{x^{2}} d x = \int x^{5} e^{u} \frac{d u}{2 x} = \frac{1}{2} \int x^{4} e^{u} d u = \frac{1}{2} \int u^{2} e^{u} d u B y T a b u l a r m e t h o d \begin{matrix} d v & w & s i g n e \\ e^{u} & u^{2} & + \\ e^{u} & 2 u & - \\ e^{u} & 2 & + \end{matrix} \frac{1}{2} \int u^{2} e^{u} d u = \frac{1}{2} u^{2} e^{u} - u e^{u} + 2 \int e^{u} d u = \frac{1}{2} u^{2} e^{u} - u e^{u} + 2 e^{u} + c b u t u = x^{2} : \int x^{5} e^{x^{2}} d x = \frac{1}{2} x^{4} e^{x^{2}} - x^{2} e^{x^{2}} + 2 e^{x^{2}} + c$

تمارين ومسائل

أجد كلاً من التكاملات الآتية:

$1 \int (x + 1) c o s x d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ c o s x & x + 1 & + \\ s i n x & 1 & - \\ - c o s x & 0 & + \end{matrix} \int (x + 1) c o s x d x = (x + 1) \sin x + \cos x + c$

$2 \int x e^{\frac{x}{2}} d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ e^{\frac{x}{2}} & x & + \\ 2 e^{\frac{x}{2}} & 1 & - \\ 4 e^{\frac{x}{2}} & 0 & + \end{matrix} \int x e^{\frac{x}{2}} d x = 2 x e^{\frac{x}{2}} - 4 e^{\frac{x}{2}} + c$

$3 \int (2 x^{2} - 1) e^{- x} d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ e^{- x} & 2 x^{2} - 1 & + \\ - e^{- x} & 4 x & - \\ e^{- x} & 4 & + \\ - e^{- x} & 0 & - \end{matrix} \int (2 x^{2} - 1) e^{- x} d x = - (2 x^{2} - 1) e^{- x} - 4 x e^{- x} - 4 e^{- x} + c$

$4 \int x l n (1 + x) d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ x & l n (1 + x) & + \\ \frac{x^{2}}{2} & \frac{1}{1 + x} & - \end{matrix} \int x l n (1 + x) d x = \frac{x^{2} l n (1 + x)}{2} - \frac{1}{2} \int \frac{x^{2}}{1 + x} d x = \frac{x^{2} l n (1 + x)}{2} - \frac{1}{2} \int (x - 1 + \frac{1}{1 + x}) d x = \frac{x^{2} l n (1 + x)}{2} - \frac{1}{2} (\frac{x^{2}}{2} - x + \ln (1 + x)) + c = \frac{x^{2} l n (1 + x)}{2} - \frac{x^{2}}{4} + \frac{x}{2} - \frac{1}{2} l n (1 + x) + c$

$5 \int x s i n x c o s x d x S o l u t i o n : B y T a b u l a r m e t h o d \int x s i n x c o s x d x = \frac{1}{2} \int 2 x s i n x c o s x d x = \frac{1}{2} \int x s i n 2 x d x \begin{matrix} d v & u & s i g n e \\ s i n 2 x & x & + \\ \frac{- c o s 2 x}{2} & 1 & - \\ \frac{- s i n 2 x}{4} & 0 & + \end{matrix} \frac{1}{2} \int x s i n 2 x d x = - \frac{x c o s 2 x}{4} + \frac{s i n 2 x}{8} + c$

$6 \int x s e c x t a n x d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ s e c x t a n x & x & + \\ s e c x & 1 & - \\ l n (s e c x + t a n x) & 0 & + \end{matrix} \int x s e c x t a n x = x s e c x - l n (s e c x + t a n x) + c O b s e r v e t h a t : \int s e c x d x = \int s e c x \times \frac{s e c x + t a n x}{s e c x + t a n x} d x = \int \frac{s e c^{2} x + s e c x t a n x}{s e c x + t a n x} d x = l n (s e c x + t a n x)$

$7 \int \frac{x}{s i n^{2} x} d x S o l u t i o n : \int \frac{x}{s i n^{2} x} d x = \int x c s c^{2} x d x B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ c s c^{2} x & x & + \\ - c o t x & 1 & - \\ - l n (s i n x) & 0 & + \end{matrix} \int \frac{x}{s i n^{2} x} d x = - x c o t x + l n (s i n x) + c$

$8 \int \frac{l n x}{x^{3}} d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ x^{- 3} & l n x & + \\ - \frac{x^{- 2}}{2} & \frac{1}{x} & - \end{matrix} \int \frac{l n x}{x^{3}} d x = - \frac{l n x}{2 x^{2}} + \frac{1}{2} \int x^{- 3} d x \int \frac{l n x}{x^{3}} d x = - \frac{l n x}{2 x^{2}} - \frac{1}{4 x^{2}} + c = - \frac{1 + l n x^{2}}{4 x^{2}} + c$

$9 \int 2 x^{2} s e c^{2} x t a n x d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ s e c^{2} x t a n x & 2 x^{2} & + \\ \frac{t a n^{2} x}{2} & 4 x & - \\ \frac{t a n x - x}{2} & 4 & + \\ - \frac{1}{2} l n (c o s x) - \frac{x^{2}}{4} & 0 & - \end{matrix} B u t \int \frac{t a n^{2} x}{2} d x = \int \frac{s e c^{2} x - 1}{2} d x = \frac{t a n x - x}{2} \int 2 x^{2} s e c^{2} x t a n x d x = x^{2} t a n^{2} x - 2 x t a n x + 2 x^{2} - 2 l n (c o s x) - x^{2} + c \int 2 x^{2} s e c^{2} x t a n x d x = x^{2} t a n^{2} x - 2 x t a n x + x^{2} - 2 l n (c o s x) + c$

$10 \int (x - 2) \sqrt{8 - x} d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ {(8 - x)}^{\frac{1}{2}} & x - 2 & + \\ - \frac{2}{3} {(8 - x)}^{\frac{3}{2}} & 1 & - \\ \frac{4}{15} {(8 - x)}^{\frac{5}{2}} & 0 & + \end{matrix} \int (x - 2) \sqrt{8 - x} d x = - \frac{2}{3} (x - 2) {(8 - x)}^{\frac{3}{2}} - \frac{4}{15} {(8 - x)}^{\frac{5}{2}} + c$

$11 \int x^{3} c o s 2 x d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ c o s 2 x & x^{3} & + \\ \frac{s i n 2 x}{2} & 3 x^{2} & - \\ - \frac{c o s 2 x}{4} & 6 x & + \\ - \frac{s i n 2 x}{8} & 6 & - \\ \frac{c o s 2 x}{16} & 0 & + \end{matrix} \int x^{3} c o s 2 x d x = \frac{x^{3} s i n 2 x}{2} + \frac{3 x^{2} c o s 2 x}{4} - \frac{6 x s i n 2 x}{8} - \frac{6 x c o s 2 x}{16} + c = \frac{x^{3} s i n 2 x}{2} + \frac{3 x^{2} c o s 2 x}{4} - \frac{3 x s i n 2 x}{4} - \frac{3 x c o s 2 x}{8} + c$

$12 \int \frac{x}{6^{x}} d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ 6^{- x} & x & + \\ \frac{- 6^{- x}}{l n 6} & 1 & - \\ \frac{6^{- x}}{{(l n 6)}^{2}} & 0 & + \end{matrix} \int \frac{x}{6^{x}} d x = - \frac{x 6^{- x}}{l n 6} - \frac{6^{- x}}{{(l n 6)}^{2}} + c = - \frac{x}{6^{x} l n 6} - \frac{1}{6^{x} {(l n 6)}^{2}} + c$

$13 \int e^{3 x} \cos x d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ e^{3 x} & \cos x & + \\ 3 e^{3 x} & - \sin x & - \\ 9 e^{3 x} & - \cos x & + \end{matrix} \int e^{3 x} \cos x d x = 3 e^{3 x} \cos x + 9 e^{3 x} \sin x - 9 \int e^{3 x} \cos x d x 10 \int e^{3 x} \cos x d x = 3 e^{3 x} \cos x + 9 e^{3 x} s i n x \int e^{3 x} c o s x d x = \frac{3 e^{3 x} \cos x + 9 e^{3 x} s i n x}{10} + c$

$14 \int c o s x l n (s i n x) d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ c o s x & l n (s i n x) & + \\ s i n x & \frac{c o s x}{s i n x} & - \end{matrix} \int c o s x l n (s i n x) d x = s i n x l n (s i n x) - \int c o s x d x \int c o s x l n (s i n x) d x = s i n x l n (s i n x) - s i n x + c$

$15 \int e^{x} l n (e^{x} + 1) d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ e^{x} & l n (e^{x} + 1) & + \\ e^{x} & \frac{e^{x}}{e^{x} + 1} & - \end{matrix} \int e^{x} l n (e^{x} + 1) d x = e^{x} l n (e^{x} + 1) - \int \frac{{(e^{x})}^{2}}{e^{x} + 1} d x H i n i t : \frac{{(e^{x})}^{2}}{e^{x} + 1} = e^{x} - \frac{e^{x}}{e^{x} + 1} = e^{x} l n (e^{x} + 1) - \int e^{x} d x + \int \frac{e^{x}}{e^{x} + 1} d x = e^{x} l n (e^{x} + 1) - e^{x} + l n (e^{x} + 1) + c = (e^{x} + 1) l n (e^{x} + 1) - e^{x} + c$

أجد كلاً من التكاملات الآتية:

$16 \int_{0}^{\frac{π}{2}} e^{x} c o s x d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ e^{x} & c o s x & + \\ e^{x} & - s i n x & - \\ e^{x} & - c o s x & + \end{matrix} \int_{0}^{\frac{π}{2}} e^{x} c o s x d x = (e^{x} c o s x + e^{x} s i n x) \frac{π}{2}^{0} - \int_{0}^{\frac{π}{2}} e^{x} c o s x d x 2 \int_{0}^{\frac{π}{2}} e^{x} c o s x d x = (e^{x} c o s x + e^{x} s i n x) \frac{π}{2}^{0} \int_{0}^{\frac{π}{2}} e^{x} c o s x d x = \frac{(e^{x} c o s x + e^{x} s i n x)}{2} \frac{π}{2}^{0} \int_{0}^{\frac{π}{2}} e^{x} c o s x d x = \frac{(e^{\frac{π}{2}} c o s (\frac{π}{2}) + e^{\frac{π}{2}} s i n (\frac{π}{2})) - (e^{0} c o s (0) + e^{0} s i n (0))}{2} = \frac{e^{\frac{π}{2}} - 1}{2}$

$17 \int_{1}^{e} l n x^{2} d x S o l u t i o n : B y T a b u l a r m e t h o d \int_{1}^{e} l n x^{2} d x = \int_{1}^{e} 2 l n x d x \begin{matrix} d v & u & s i g n e \\ d x & 2 \ln x & + \\ x & \frac{2}{x} & - \end{matrix} \int_{1}^{e} l n x^{2} d x = 2 x \ln x e^{1} - \int_{1}^{e} 2 d x = 2 e l n e - 2 \ln 1 - 2 (e - 1) = 2$

$18 \int_{1}^{2} l n (x e^{x}) d x S o l u t i o n : B y T a b u l a r m e t h o d \int_{1}^{2} l n (x e^{x}) d x = \int_{1}^{2} l n x d x + \int_{1}^{2} l n (e^{x}) d x = \int_{1}^{2} l n x d x + \int_{1}^{2} x d x \begin{matrix} d v & u & s i g n e \\ d x & l n x & + \\ x & \frac{1}{x} & - \end{matrix} \int_{1}^{2} l n (x e^{x}) d x = x l n x 2^{1} - \int_{1}^{2} 1 d x + \int_{1}^{2} x d x = 2 l n 2 - l n 1 - 1 (2 - 1) + \frac{x^{2}}{2} 2^{1} = l n 4 + \frac{1}{2}$

$19 \int_{\frac{π}{12}}^{\frac{π}{9}} x s e c^{2} 3 x d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ s e c^{2} 3 x & x & + \\ \frac{\tan 3 x}{3} & 1 & - \\ - \frac{1}{9} \ln (\cos 3 x) & 0 & + \end{matrix} \int_{\frac{π}{12}}^{\frac{π}{9}} x s e c^{2} 3 x d x = \frac{x t a n 3 x}{3} + \frac{1}{9} l n (c o s 3 x) \frac{π}{9}^{\frac{π}{12}} = \frac{\frac{π}{9} \tan (\frac{π}{3}) - \frac{π}{12} t a n (\frac{π}{4})}{3} + \frac{1}{9} l n (c o s (\frac{π}{3})) - \frac{1}{9} l n (c o s (\frac{π}{4})) = \frac{\frac{\sqrt{3} π}{9} - \frac{π}{12}}{3} + \frac{1}{9} (\ln (\frac{1}{2}) - \ln (\frac{1}{\sqrt{2}})) = \frac{π (4 \sqrt{3} - 3)}{108} + \frac{1}{9} (- l n 2 - \frac{1}{2} l n 2) = \frac{π (4 \sqrt{3} - 3)}{108} + \frac{1}{9} (- \frac{3}{2} l n 2) = \frac{π (4 \sqrt{3} - 3)}{108} - \frac{1}{6} l n 2$

$20 \int_{1}^{e} x^{4} l n x d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ x^{4} & l n x & + \\ \frac{x^{5}}{5} & \frac{1}{x} & - \end{matrix} \int_{1}^{e} x^{4} l n x d x = \frac{x^{5}}{5} l n x e^{1} - \frac{1}{5} \int_{1}^{e} x^{4} d x = \frac{x^{5}}{5} l n x e^{1} - \frac{x^{5}}{25} e^{1} = \frac{e^{5} \ln e}{5} - \frac{l n 1}{5} - \frac{e^{5} - 1}{25} = \frac{e^{5}}{5} - \frac{e^{5} - 1}{25} = \frac{4 e^{5} + 1}{25}$

$21 \int_{0}^{\frac{π}{2}} x^{2} s i n x d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ s i n x & x^{2} & + \\ - c o s x & 2 x & - \\ - s i n x & 2 & + \\ c o s x & 0 & - \end{matrix} \int_{0}^{\frac{π}{2}} x^{2} s i n x d x = (- x^{2} c o s x + 2 x s i n x + 2 c o s x) \frac{π}{2}^{0} = (- {(\frac{π}{2})}^{2} c o s (\frac{π}{2}) + 2 (\frac{π}{2}) s i n (\frac{π}{2}) + 2 c o s (\frac{π}{2})) - (- {(0)}^{2} c o s (0) + 2 (0) s i n (0) + 2 c o s (0)) = π - 2$

$22 \int_{0}^{1} x (e^{- 2 x} + e^{- x}) d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ e^{- 2 x} + e^{- x} & x & + \\ \frac{e^{- 2 x}}{- 2} - e^{- x} & 1 & - \\ \frac{e^{- 2 x}}{4} + e^{- x} & 0 & + \end{matrix} \int_{0}^{1} x (e^{- 2 x} + e^{- x}) d x = x (\frac{e^{- 2 x}}{- 2} - e^{- x}) - (\frac{e^{- 2 x}}{4} + e^{- x}) 1 0 = (\frac{e^{- 2}}{- 2} - e^{- 1}) - (\frac{e^{- 2}}{4} + e^{- 1}) - 0 (\frac{e^{0}}{- 2} - e^{0}) + (\frac{e^{0}}{4} + e^{0}) = (\frac{e^{- 2}}{- 2} - e^{- 1}) - (\frac{e^{- 2}}{4} + e^{- 1}) - 0 + (\frac{e^{0}}{4} + e^{0}) = (\frac{e^{- 2}}{- 2} - e^{- 1}) - (\frac{e^{- 2}}{4} + e^{- 1}) + (\frac{e^{0}}{4} + e^{0}) = \frac{- 3}{4 e^{2}} - \frac{2}{e} + \frac{5}{4}$

$23 \int_{0}^{1} \frac{x e^{x}}{{(1 + x)}^{2}} d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ {(1 + x)}^{- 2} & x e^{x} & + \\ \frac{- 1}{1 + x} & e^{x} (1 + x) & - \end{matrix} \int_{0}^{1} \frac{x e^{x}}{{(1 + x)}^{2}} d x = - \frac{x e^{x}}{1 + x} 1^{0} + \int_{0}^{1} \frac{e^{x} (1 + x)}{(1 + x)} d x = - \frac{e}{2} + \int_{0}^{1} e^{x} d x = - \frac{e}{2} + e - 1 = \frac{e}{2} - 1$

$24 \int_{0}^{1} x 3^{x} d x S o l u t i o n : B y T a b u l a r m e t h o d \begin{matrix} d v & u & s i g n e \\ 3^{x} & x & + \\ \frac{3^{x}}{l n 3} & 1 & - \\ \frac{3^{x}}{{(l n 3)}^{2}} & 0 & + \end{matrix} \int_{0}^{1} x 3^{x} d x = \frac{x 3^{x}}{l n 3} - \frac{3^{x}}{{(l n 3)}^{2}} 1 0 = (\frac{3}{l n 3} - \frac{3}{{(l n 3)}^{2}}) - (- \frac{1}{{(l n 3)}^{2}})$