Trigonometric substitution Half-Angle Substitution Integration of Rational Functions by Partial Fractions # Trigonometric substitution含有 a 2 − x 2 \sqrt{a^2-x^2} a 2 − x 2 a 2 + x 2 \sqrt{a^2+x^2} a 2 + x 2 x 2 − a 2 \sqrt{x^2-a^2} x 2 − a 2
a 2 − x 2 \sqrt{a^2-x^2} a 2 − x 2 x = a s i n θ , − π 2 < θ < π 2 x=asin \theta ,\; -\frac{\pi}{2}< \theta < \frac{\pi}{2} x = a s i n θ , − 2 π < θ < 2 π a 2 − x 2 = a c o s θ \sqrt{a^2-x^2} = acos\theta a 2 − x 2 = a c o s θ a 2 + x 2 \sqrt{a^2+x^2} a 2 + x 2 x = a t a n θ , − π 2 < θ < π 2 x=atan \theta ,\; -\frac{\pi}{2}< \theta < \frac{\pi}{2} x = a t a n θ , − 2 π < θ < 2 π a 2 + x 2 = a s e c θ \sqrt{a^2+x^2} = asec\theta a 2 + x 2 = a s e c θ x 2 − a 2 \sqrt{x^2-a^2} x 2 − a 2 x = a s e c θ , 0 ≤ θ < π 2 o r π ≤ θ < 3 2 π x=asec \theta ,\; 0 \le \theta < \frac{\pi}{2} \; or \; \pi \le \theta < \frac{3}{2}\pi x = a s e c θ , 0 ≤ θ < 2 π o r π ≤ θ < 2 3 π x 2 − a 2 = a t a n θ \sqrt{x^2-a^2} = atan\theta x 2 − a 2 = a t a n θ ∫ 9 − x 2 x 2 d x \int \frac{\sqrt{9-x^2}}{x^2}dx ∫ x 2 9 − x 2 d x ∫ b a a 2 − x 2 d x \int \frac{b}{a}\sqrt{a^2-x^2}dx ∫ a b a 2 − x 2 d x ∫ 1 x 2 x 2 + 4 d x \int \frac{1}{x^2\sqrt{x^2+4}}dx ∫ x 2 x 2 + 4 1 d x ∫ 1 x 2 − a 2 d x \int \frac{1}{\sqrt{x^2-a^2}}dx ∫ x 2 − a 2 1 d x ∫ x 3 ( 4 x 2 + 9 ) 3 2 d x \int \frac{x^3}{(4x^2+9)^{\frac{3}{2}}}dx ∫ ( 4 x 2 + 9 ) 2 3 x 3 d x ∫ x 3 − 2 x − x 2 d x \int \frac{x}{\sqrt{3-2x-x^2}}dx ∫ 3 − 2 x − x 2 x d x Let x = 3 s i n θ , − π 2 ≤ θ ≤ π 2 ⇒ d x = 3 c o s θ d θ x = 3sin\theta ,-\frac{\pi}{2} \le \theta \le \frac{\pi}{2} \Rightarrow dx = 3cos\theta d \theta x = 3 s i n θ , − 2 π ≤ θ ≤ 2 π ⇒ d x = 3 c o s θ d θ
∫ 9 − x 2 x 2 d x = ∫ 3 c o s θ 9 s i n 2 θ ⋅ 3 c o s θ d θ = ∫ c o t 2 θ d θ = ∫ ( c s c 2 θ − 1 ) d θ = − c o t θ − θ + k = − 9 − x 2 x − s i n − 1 ( x 3 ) + k \int \frac{\sqrt{9-x^2}}{x^2}dx = \int \frac{3cos\theta}{9sin^2\theta} \cdot 3cos\theta d \theta = \int cot^2\theta d \theta = \int (csc^2\theta -1) d \theta \\ = -cot \theta - \theta + k = -\frac{\sqrt{9-x^2}}{x} - sin^{-1}(\frac{x}{3})+k ∫ x 2 9 − x 2 d x = ∫ 9 s i n 2 θ 3 c o s θ ⋅ 3 c o s θ d θ = ∫ c o t 2 θ d θ = ∫ ( c s c 2 θ − 1 ) d θ = − c o t θ − θ + k = − x 9 − x 2 − s i n − 1 ( 3 x ) + k
Let x = a s i n θ , − π 2 ≤ θ ≤ π 2 ⇒ d x = a c o s θ d θ x = asin\theta, -\frac{\pi}{2} \le \theta \le \frac{\pi}{2} \Rightarrow dx = acos\theta d\theta x = a s i n θ , − 2 π ≤ θ ≤ 2 π ⇒ d x = a c o s θ d θ ∫ b a a 2 − x 2 d x = b a ∫ a c o s θ a c o s θ d θ = a b ∫ c o s 2 θ d θ = a b 2 ∫ ( 1 + c o s 2 θ ) d θ = a b 2 ( θ + 1 2 s i n 2 θ ) + k = a b 2 ( s i n − 1 ( x a + x a 2 a 2 − x 2 ) + k \int \frac{b}{a}\sqrt{a^2-x^2}dx = \frac{b}{a}\int acos\theta acos\theta d\theta = ab \int cos^2\theta d\theta = \frac{ab}{2}\int (1+cos2\theta)d\theta = \frac{ab}{2}(\theta + \frac{1}{2}sin2\theta) + k = \frac{ab}{2}(sin^{-1}(\frac{x}{a} + \frac{x}{a^2}\sqrt{a^2-x^2})+k ∫ a b a 2 − x 2 d x = a b ∫ a c o s θ a c o s θ d θ = a b ∫ c o s 2 θ d θ = 2 a b ∫ ( 1 + c o s 2 θ ) d θ = 2 a b ( θ + 2 1 s i n 2 θ ) + k = 2 a b ( s i n − 1 ( a x + a 2 x a 2 − x 2 ) + k
Let x = 2 t a n θ ⇒ d x = 2 s e c 2 θ d θ x = 2tan \theta \Rightarrow dx = 2sec^2\theta d \theta x = 2 t a n θ ⇒ d x = 2 s e c 2 θ d θ ∴ ∫ 1 x 2 x 2 + 4 d x = ∫ 1 4 t a n 2 θ ⋅ 2 s e c θ ( 2 s e c 2 θ d θ ) = 1 4 ∫ s e c θ t a n 2 θ d θ = 1 4 c o s θ s i n 2 θ d θ = 1 4 ∫ ( s i n θ ) − 2 d ( s i n θ ) = − 1 4 s i n − 1 θ + k = − 1 4 x 2 + 4 x + k \therefore \int \frac{1}{x^2\sqrt{x^2+4}}dx = \int \frac{1}{4tan^2\theta \cdot 2sec\theta}(2sec^2\theta d\theta) = \frac{1}{4}\int \frac{sec\theta}{tan^2\theta}d\theta = \frac{1}{4}\frac{cos\theta}{sin^2\theta} d\theta= \frac{1}{4}\int (sin\theta)^{-2}d(sin\theta) = -\frac{1}{4}sin^{-1}\theta+k = -\frac{1}{4}\frac{\sqrt{x^2+4}}{x}+k ∴ ∫ x 2 x 2 + 4 1 d x = ∫ 4 t a n 2 θ ⋅ 2 s e c θ 1 ( 2 s e c 2 θ d θ ) = 4 1 ∫ t a n 2 θ s e c θ d θ = 4 1 s i n 2 θ c o s θ d θ = 4 1 ∫ ( s i n θ ) − 2 d ( s i n θ ) = − 4 1 s i n − 1 θ + k = − 4 1 x x 2 + 4 + k
Let x = a s e c θ ⇒ d x = a t a n θ s e c θ d θ x = asec\theta \Rightarrow dx = atan\theta sec\theta d\theta x = a s e c θ ⇒ d x = a t a n θ s e c θ d θ ∫ 1 x 2 − a 2 d x = ∫ 1 a t a n θ a t a n θ s e c θ d θ = ∫ s e c θ d θ = l n ∣ s e c θ + t a n θ ∣ + k = l n ∣ x a + x 2 − a 2 a ∣ + k \int \frac{1}{\sqrt{x^2-a^2}}dx = \int \frac{1}{atan\theta}atan\theta sec\theta d\theta = \int sec\theta d\theta = ln|sec\theta + tan\theta|+k = ln|\frac{x}{a}+\frac{\sqrt{x^2-a^2}}{a}|+k ∫ x 2 − a 2 1 d x = ∫ a t a n θ 1 a t a n θ s e c θ d θ = ∫ s e c θ d θ = l n ∣ s e c θ + t a n θ ∣ + k = l n ∣ a x + a x 2 − a 2 ∣ + k = l n ∣ x + x 2 − a 2 ∣ − l n a + k = l n ∣ x + x 2 − a 2 ∣ + k = ln|x+\sqrt{x^2-a^2}|-lna+k = ln|x+\sqrt{x^2-a^2}|+k = l n ∣ x + x 2 − a 2 ∣ − l n a + k = l n ∣ x + x 2 − a 2 ∣ + k
Let x = 3 2 t a n θ ⇒ d x = 3 2 s e c 2 θ d θ x = \frac{3}{2}tan\theta \Rightarrow dx = \frac{3}{2}sec^2\theta d\theta x = 2 3 t a n θ ⇒ d x = 2 3 s e c 2 θ d θ ∫ x 3 ( 4 x 2 + 9 ) 3 2 d x = ∫ 27 8 t a n 3 θ 27 s e c 3 θ ⋅ 3 2 s e c 2 θ d θ = 3 16 ∫ t a n 3 θ s e c θ d θ = 3 16 ∫ s i n 3 θ c o s 2 θ d θ = 3 16 ∫ s i n 2 θ c o s 2 θ d θ = 3 16 ∫ 1 − c o s 2 θ c o s 2 θ d ( c o s θ ) = 3 16 ∫ ( 1 − ( c o s θ ) − 2 ) d ( c o s θ ) = 3 16 ( c o s θ + 1 c o s θ ) + k = 3 16 ( 3 4 x 2 + 9 + 4 x 2 + 9 3 ) + k \int \frac{x^3}{(4x^2+9)^{\frac{3}{2}}}dx = \int \frac{\frac{27}{8}tan^3\theta}{27sec^3\theta} \cdot \frac{3}{2}sec^2\theta d\theta = \frac{3}{16}\int \frac{tan^3\theta}{sec\theta}d\theta = \frac{3}{16}\int \frac{sin^3\theta}{cos^2\theta}d\theta \\= \frac{3}{16}\int \frac{sin^2\theta}{cos^2\theta}d\theta = \frac{3}{16}\int \frac{1-cos^2\theta}{cos^2\theta}d(cos\theta) = \frac{3}{16}\int (1-(cos\theta)^{-2})d(cos\theta) \\= \frac{3}{16}(cos\theta + \frac{1}{cos\theta})+k = \frac{3}{16}(\frac{3}{\sqrt{4x^2+9}} + \frac{\sqrt{4x^2+9}}{3})+k ∫ ( 4 x 2 + 9 ) 2 3 x 3 d x = ∫ 2 7 s e c 3 θ 8 2 7 t a n 3 θ ⋅ 2 3 s e c 2 θ d θ = 1 6 3 ∫ s e c θ t a n 3 θ d θ = 1 6 3 ∫ c o s 2 θ s i n 3 θ d θ = 1 6 3 ∫ c o s 2 θ s i n 2 θ d θ = 1 6 3 ∫ c o s 2 θ 1 − c o s 2 θ d ( c o s θ ) = 1 6 3 ∫ ( 1 − ( c o s θ ) − 2 ) d ( c o s θ ) = 1 6 3 ( c o s θ + c o s θ 1 ) + k = 1 6 3 ( 4 x 2 + 9 3 + 3 4 x 2 + 9 ) + k
Let x + 1 = 2 s i n θ ⇒ d x = 2 c o s θ d θ x+1 = 2sin\theta \Rightarrow dx = 2cos\theta d\theta x + 1 = 2 s i n θ ⇒ d x = 2 c o s θ d θ ∫ x 3 − 2 x − x 2 d x = ∫ 2 s i n θ − 1 2 c o s θ 2 c o s θ d θ = ∫ ( 2 s i n θ − 1 ) d θ = − 2 c o s θ − θ + k = − 3 − 2 x − x 2 − s i n − 1 ( x + 1 2 + k \int \frac{x}{\sqrt{3-2x-x^2}}dx = \int \frac{2sin\theta -1}{2cos\theta} 2cos\theta d\theta = \int (2sin\theta -1)d\theta \\= -2cos\theta - \theta +k = -\sqrt{3-2x-x^2} - sin^{-1}(\frac{x+1}{2}+k ∫ 3 − 2 x − x 2 x d x = ∫ 2 c o s θ 2 s i n θ − 1 2 c o s θ d θ = ∫ ( 2 s i n θ − 1 ) d θ = − 2 c o s θ − θ + k = − 3 − 2 x − x 2 − s i n − 1 ( 2 x + 1 + k
# Half-Angle SubstitutionLet t = t a n x 2 t = tan\frac{x}{2} t = t a n 2 x
s i n x 2 = t 1 + t 2 sin\frac{x}{2} = \frac{t}{\sqrt{1+t^2}} s i n 2 x = 1 + t 2 t c o s x 2 = 1 1 + t 2 cos\frac{x}{2} = \frac{1}{\sqrt{1+t^2}} c o s 2 x = 1 + t 2 1 s i n x = 2 s i n x 2 c o s x 2 = 2 t 1 + t 2 sinx = 2sin\frac{x}{2}cos\frac{x}{2} = \frac{2t}{1+t^2} s i n x = 2 s i n 2 x c o s 2 x = 1 + t 2 2 t c o s x = 2 c o s 2 x 2 − 1 = 1 − t 2 1 + t 2 cosx = 2cos^2\frac{x}{2} -1 = \frac{1-t^2}{1+t^2} c o s x = 2 c o s 2 2 x − 1 = 1 + t 2 1 − t 2 d t = 1 2 s e c 2 x 2 d x = 1 + t 2 2 d x dt = \frac{1}{2}sec^2\frac{x}{2}dx = \frac{1+t^2}{2}dx d t = 2 1 s e c 2 2 x d x = 2 1 + t 2 d x ∴ d x = 2 1 + t 2 d t \therefore dx = \frac{2}{1+t^2}dt ∴ d x = 1 + t 2 2 d t ∫ 1 1 + s i n x + c o s x d x \int \frac{1}{1+sinx+cosx}dx ∫ 1 + s i n x + c o s x 1 d x ∫ 1 1 + s i n x − c o s x d x \int \frac{1}{1+sinx-cosx}dx ∫ 1 + s i n x − c o s x 1 d x Let t = t a n x 2 ⇒ s i n x = 2 t 1 + t 2 , c o s x = 1 − t 2 1 + t 2 , d x = 2 1 + t 2 d t t = tan\frac{x}{2} \Rightarrow sinx = \frac{2t}{1+t^2}, cosx = \frac{1-t^2}{1+t^2},dx = \frac{2}{1+t^2}dt t = t a n 2 x ⇒ s i n x = 1 + t 2 2 t , c o s x = 1 + t 2 1 − t 2 , d x = 1 + t 2 2 d t ∫ 1 1 + s i n x + c o s x d x = ∫ 1 1 + 2 t 1 + t 2 + 1 − t 2 1 + t 2 2 1 + t 2 d t = ∫ 2 1 + t 2 + 2 t + 1 − t 2 d t = ∫ 1 t + 1 d t = l n ∣ t + 1 ∣ + k = l n ∣ t a n x 2 + 1 ∣ + k \int \frac{1}{1+sinx+cosx}dx = \int \frac{1}{1+\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}} \frac{2}{1+t^2}dt = \int \frac{2}{1+t^2+2t+1-t^2} dt \\= \int \frac{1}{t+1}dt = ln|t+1|+k = ln|tan\frac{x}{2}+1|+k ∫ 1 + s i n x + c o s x 1 d x = ∫ 1 + 1 + t 2 2 t + 1 + t 2 1 − t 2 1 1 + t 2 2 d t = ∫ 1 + t 2 + 2 t + 1 − t 2 2 d t = ∫ t + 1 1 d t = l n ∣ t + 1 ∣ + k = l n ∣ t a n 2 x + 1 ∣ + k
Let t = t a n x 2 ⇒ s i n x = 2 t 1 + t 2 , c o s x = 1 − t 2 1 + t 2 , d x = 2 1 + t 2 d t t = tan\frac{x}{2} \Rightarrow sinx = \frac{2t}{1+t^2}, cosx = \frac{1-t^2}{1+t^2},dx = \frac{2}{1+t^2}dt t = t a n 2 x ⇒ s i n x = 1 + t 2 2 t , c o s x = 1 + t 2 1 − t 2 , d x = 1 + t 2 2 d t ∫ 1 1 + s i n x − c o s x d x = ∫ 1 1 + 2 t 1 + t 2 − 1 − t 2 1 + t 2 2 1 + t 2 d t = ∫ 2 1 + t 2 + 2 t − 1 + t 2 d t = ∫ 1 t ( t + 1 ) d t = ∫ ( 1 t + − 1 t + 1 ) d t = l n ∣ t ∣ − l n ∣ t + 1 ∣ + k = l n ∣ t a n x 2 ∣ − l n ∣ t a n x 2 + 1 ∣ + k \int \frac{1}{1+sinx-cosx}dx = \int \frac{1}{1+\frac{2t}{1+t^2}-\frac{1-t^2}{1+t^2}} \frac{2}{1+t^2}dt = \int \frac{2}{1+t^2+2t-1+t^2} dt = \int \frac{1}{t(t+1)}dt = \int (\frac{1}{t}+\frac{-1}{t+1}) dt \\= ln|t| - ln|t+1|+k = ln|tan\frac{x}{2}| - ln|tan\frac{x}{2}+1|+k ∫ 1 + s i n x − c o s x 1 d x = ∫ 1 + 1 + t 2 2 t − 1 + t 2 1 − t 2 1 1 + t 2 2 d t = ∫ 1 + t 2 + 2 t − 1 + t 2 2 d t = ∫ t ( t + 1 ) 1 d t = ∫ ( t 1 + t + 1 − 1 ) d t = l n ∣ t ∣ − l n ∣ t + 1 ∣ + k = l n ∣ t a n 2 x ∣ − l n ∣ t a n 2 x + 1 ∣ + k
∫ 1 3 + c o s x d x \int \frac{1}{3+cosx} dx ∫ 3 + c o s x 1 d x
Let t = t a n x 2 ⇒ c o s x = 1 − t 2 1 + t 2 , d x = 2 1 + t 2 d t t = tan\frac{x}{2} \Rightarrow cosx = \frac{1-t^2}{1+t^2},dx = \frac{2}{1+t^2}dt t = t a n 2 x ⇒ c o s x = 1 + t 2 1 − t 2 , d x = 1 + t 2 2 d t ∫ 1 3 + c o s x d x = ∫ 1 3 + 1 − t 2 1 + t 2 2 1 + t 2 d t = ∫ 2 3 + t 2 + 1 − t 2 d t = ∫ 1 2 + t 2 d t \int \frac{1}{3+cosx} dx = \int \frac{1}{3+\frac{1-t^2}{1+t^2}} \frac{2}{1+t^2}dt = \int \frac{2}{3+t^2+1-t^2}dt = \int \frac{1}{2+t^2}dt ∫ 3 + c o s x 1 d x = ∫ 3 + 1 + t 2 1 − t 2 1 1 + t 2 2 d t = ∫ 3 + t 2 + 1 − t 2 2 d t = ∫ 2 + t 2 1 d t t = 2 t a n θ ⇒ d t = 2 s e c 2 θ d θ , θ = t a n − 1 ( 1 2 t ) t = \sqrt{2}tan\theta \Rightarrow dt = \sqrt{2}sec^2\theta d\theta , \theta = tan^{-1}(\frac{1}{\sqrt{2}}t) t = 2 t a n θ ⇒ d t = 2 s e c 2 θ d θ , θ = t a n − 1 ( 2 1 t ) ∫ 1 2 + t 2 d t = ∫ 1 2 + 2 t a n θ 2 s e c 2 θ d θ = ∫ 2 s e c 2 θ 2 s e c 2 θ d θ = ∫ 1 2 d θ = 1 2 θ + k = 1 2 t a n − 1 ( t 2 ) + k = 1 2 t a n − 1 ( t a n x 2 2 ) + k \int \frac{1}{2+t^2}dt = \int \frac{1}{2+2tan\theta}\sqrt{2}sec^2\theta d\theta = \int \frac{\sqrt{2}sec^2\theta}{2sec^2\theta}d\theta = \int \frac{1}{\sqrt{2}}d\theta = \frac{1}{\sqrt{2}}\theta+k \\= \frac{1}{\sqrt{2}}tan^{-1}(\frac{t}{\sqrt{2}})+k = \frac{1}{\sqrt{2}}tan^{-1}(\frac{tan\frac{x}{2}}{\sqrt{2}})+k ∫ 2 + t 2 1 d t = ∫ 2 + 2 t a n θ 1 2 s e c 2 θ d θ = ∫ 2 s e c 2 θ 2 s e c 2 θ d θ = ∫ 2 1 d θ = 2 1 θ + k = 2 1 t a n − 1 ( 2 t ) + k = 2 1 t a n − 1 ( 2 t a n 2 x ) + k
# Integration of Rational Functions by Partial FractionsCase 1 ∫ P ( x ) Q ( x ) d x = ∫ M ( x ) d x + ∫ r ( x ) Q ( x ) d x \int \frac{P(x)}{Q(x)}dx = \int M(x)dx + \int \frac{r(x)}{Q(x)}dx ∫ Q ( x ) P ( x ) d x = ∫ M ( x ) d x + ∫ Q ( x ) r ( x ) d x r ( x ) Q ( x ) \frac{r(x)}{Q(x)} Q ( x ) r ( x ) Case 2 Q ( x ) Q(x) Q ( x )
不可分解，則進行配方 後，使用三角代換法 。 可分解為( a x + b ) n o r ( a x 2 + b x + c ) n (ax+b)^n \; or \;(ax^2+bx+c)^n ( a x + b ) n o r ( a x 2 + b x + c ) n Case 3
每個( a x + b ) n (ax+b)^n ( a x + b ) n A 1 a x + b + A 2 ( a x + b ) 2 + . . . + A n ( a x + b ) n \frac{A_1}{ax+b}+\frac{A_2}{(ax+b)^2}+...+\frac{A_n}{(ax+b)^n} a x + b A 1 + ( a x + b ) 2 A 2 + . . . + ( a x + b ) n A n 每個( a x 2 + b x + c ) n (ax^2+bx+c)^n ( a x 2 + b x + c ) n A 1 x + B 1 a x 2 + b x + c + A 2 x + B 2 ( a x 2 + b x + c ) 2 + . . . + A n x + B n ( a x 2 + b x + c ) n \frac{A_1x+B_1}{ax^2+bx+c}+\frac{A_2x+B_2}{(ax^2+bx+c)^2}+...+\frac{A_nx+B_n}{(ax^2+bx+c)^n} a x 2 + b x + c A 1 x + B 1 + ( a x 2 + b x + c ) 2 A 2 x + B 2 + . . . + ( a x 2 + b x + c ) n A n x + B n A i , B i f o r i ∈ { 1 , 2 , . . . , n } A_i,B_i \; for \; i\in \{ 1,2,...,n \} A i , B i f o r i ∈ { 1 , 2 , . . . , n } 三角代換，常用公式 ∫ 1 a 2 + x 2 d x = 1 a t a n − 1 ( x a ) + k \int \frac{1}{a^2+x^2}dx = \frac{1}{a}tan^{-1}(\frac{x}{a})+k ∫ a 2 + x 2 1 d x = a 1 t a n − 1 ( a x ) + k ∫ 1 a 2 − x 2 d x = s i n − 1 ( x a ) + k \int \frac{1}{\sqrt{a^2-x^2}}dx = sin^{-1}(\frac{x}{a})+k ∫ a 2 − x 2 1 d x = s i n − 1 ( a x ) + k ∫ 1 x x 2 − a 2 d x = 1 a s e c − 1 ∣ x a ∣ + k \int \frac{1}{x\sqrt{x^2-a^2}}dx = \frac{1}{a}sec^{-1}|\frac{x}{a}|+k ∫ x x 2 − a 2 1 d x = a 1 s e c − 1 ∣ a x ∣ + k Case 1、Case 2
∫ x 3 + x x − 1 d x \int \frac{x^3+x}{x-1} dx ∫ x − 1 x 3 + x d x
∫ x 3 + x x − 1 d x = ∫ ( x 2 + x + 2 + 2 x − 1 ) d x = 1 3 x 3 + 1 2 x 2 + 2 x + 2 ∫ 1 x − 1 d x = 1 3 x 3 + 1 2 x 2 + 2 x + 2 l n ∣ x − 1 ∣ + k \int \frac{x^3+x}{x-1} dx = \int (x^2+x+2 + \frac{2}{x-1})dx \\= \frac{1}{3}x^3 + \frac{1}{2}x^2+2x+2 \int \frac{1}{x-1}dx \\= \frac{1}{3}x^3 + \frac{1}{2}x^2+2x+2ln|x-1|+k ∫ x − 1 x 3 + x d x = ∫ ( x 2 + x + 2 + x − 1 2 ) d x = 3 1 x 3 + 2 1 x 2 + 2 x + 2 ∫ x − 1 1 d x = 3 1 x 3 + 2 1 x 2 + 2 x + 2 l n ∣ x − 1 ∣ + k
Case 1、Case 2、Case 3
∫ x 3 − 2 x − 4 ( x 2 − x ) ( x 2 + 4 ) d x \int \frac{x^3-2x-4}{(x^2-x)(x^2+4)}dx ∫ ( x 2 − x ) ( x 2 + 4 ) x 3 − 2 x − 4 d x
∫ x 3 − 2 x − 4 ( x 2 − x ) ( x 2 + 4 ) d x = ∫ x 3 − 2 x − 4 x ( x − 1 ) ( x 2 + 4 ) d x = ∫ ( A x + B x − 1 + C x + D x 2 + 4 ) d x \int \frac{x^3-2x-4}{(x^2-x)(x^2+4)}dx = \int \frac{x^3-2x-4}{x(x-1)(x^2+4)}dx = \int ( \frac{A}{x} + \frac{B}{x-1} + \frac{Cx+D}{x^2+4})dx ∫ ( x 2 − x ) ( x 2 + 4 ) x 3 − 2 x − 4 d x = ∫ x ( x − 1 ) ( x 2 + 4 ) x 3 − 2 x − 4 d x = ∫ ( x A + x − 1 B + x 2 + 4 C x + D ) d x ⇒ x 3 − 2 x − 4 = A ( x − 1 ) ( x 2 + 4 ) + B x ( x 2 + 4 ) + ( C x + D ) x ( x + 1 ) \Rightarrow x^3-2x-4 = A(x-1)(x^2+4)+Bx(x^2+4)+(Cx+D)x(x+1) ⇒ x 3 − 2 x − 4 = A ( x − 1 ) ( x 2 + 4 ) + B x ( x 2 + 4 ) + ( C x + D ) x ( x + 1 ) x = 0 , 1 , − 1 , 2 x = 0,1,-1,2 x = 0 , 1 , − 1 , 2 ⇒ A = 1 , B = − 1 , C = 1 , D = 2 \Rightarrow A = 1,B=-1,C=1,D=2 ⇒ A = 1 , B = − 1 , C = 1 , D = 2 ∫ ( A x + B x − 1 + C x + D x 2 + 4 ) d x = ∫ ( 1 x + − 1 x − 1 + x + 2 x 2 + 4 ) d x = l n ∣ x ∣ − l n ∣ x − 1 ∣ + 1 2 l n ∣ x 2 + 4 ∣ + t a n − 1 ( x 2 ) + k \int ( \frac{A}{x} + \frac{B}{x-1} + \frac{Cx+D}{x^2+4})dx = \int ( \frac{1}{x} + \frac{-1}{x-1} + \frac{x+2}{x^2+4})dx \\= ln|x|-ln|x-1|+\frac{1}{2}ln|x^2+4| + tan^{-1}(\frac{x}{2})+k ∫ ( x A + x − 1 B + x 2 + 4 C x + D ) d x = ∫ ( x 1 + x − 1 − 1 + x 2 + 4 x + 2 ) d x = l n ∣ x ∣ − l n ∣ x − 1 ∣ + 2 1 l n ∣ x 2 + 4 ∣ + t a n − 1 ( 2 x ) + k
Case 1、Case 2
∫ 4 x 3 − 3 x + 2 4 x 2 − 4 x + 3 d x \int \frac{4x^3-3x+2}{4x^2-4x+3}dx ∫ 4 x 2 − 4 x + 3 4 x 3 − 3 x + 2 d x
∫ 4 x 3 − 3 x + 2 4 x 2 − 4 x + 3 d x = ∫ ( x + 1 + − 2 x − 1 4 x 2 − 4 x + 3 ) d x \int \frac{4x^3-3x+2}{4x^2-4x+3}dx = \int (x+1 + \frac{-2x-1}{4x^2-4x+3})dx ∫ 4 x 2 − 4 x + 3 4 x 3 − 3 x + 2 d x = ∫ ( x + 1 + 4 x 2 − 4 x + 3 − 2 x − 1 ) d x = 1 2 x 2 + x − 2 ∫ x ( 2 x − 1 ) 2 + 2 d x − ∫ 1 ( 2 x − 1 ) 2 + 2 d x = \frac{1}{2}x^2 + x -2\int \frac{x}{(2x-1)^2+2}dx - \int \frac{1}{(2x-1)^2+2}dx = 2 1 x 2 + x − 2 ∫ ( 2 x − 1 ) 2 + 2 x d x − ∫ ( 2 x − 1 ) 2 + 2 1 d x = 1 2 x 2 + x − 1 4 l n ∣ 4 x 2 − 4 x + 3 ∣ − 1 2 t a n − 1 ( 2 x − 1 2 ) + k = \frac{1}{2}x^2 + x - \frac{1}{4}ln|4x^2-4x+3| - \frac{1}{\sqrt{2}}tan^{-1}(\frac{2x-1}{\sqrt{2}}) +k = 2 1 x 2 + x − 4 1 l n ∣ 4 x 2 − 4 x + 3 ∣ − 2 1 t a n − 1 ( 2 2 x − 1 ) + k
Let u = 2 x − 1 ⇒ d u = 2 d x u = 2x-1 \Rightarrow du = 2dx u = 2 x − 1 ⇒ d u = 2 d x
∫ 1 ( 2 x − 1 ) 2 + 2 d x = 1 2 ∫ 1 u 2 + 2 2 d u = 1 2 2 t a n − 1 ( u 2 ) + k = 1 2 2 t a n − 1 ( 2 x − 1 2 ) + k \int \frac{1}{(2x-1)^2+2}dx = \frac{1}{2}\int \frac{1}{u^2+\sqrt{2}^2}du = \frac{1}{2\sqrt{2}} tan^{-1}(\frac{u}{\sqrt{2}})+k = \frac{1}{2\sqrt{2}} tan^{-1}(\frac{2x-1}{\sqrt{2}})+k ∫ ( 2 x − 1 ) 2 + 2 1 d x = 2 1 ∫ u 2 + 2 2 1 d u = 2 2 1 t a n − 1 ( 2 u ) + k = 2 2 1 t a n − 1 ( 2 2 x − 1 ) + k ∫ x ( 2 x − 1 ) 2 + 2 d x = ∫ 1 2 ( u + 1 ) u 2 + 2 1 2 d u = 1 4 ∫ u u 2 + 2 d u + 1 4 ∫ 1 u 2 + 2 d u = 1 8 l n ∣ u 2 + 2 ∣ + 1 4 ⋅ 1 2 t a n − 1 ( u 2 ) + k = 1 8 l n ∣ 4 x 2 − 4 x + 3 ∣ + 1 4 2 t a n − 1 ( 2 x − 1 2 ) + k \int \frac{x}{(2x-1)^2+2}dx = \int \frac{\frac{1}{2}(u+1)}{u^2+2}\frac{1}{2}du = \frac{1}{4} \int \frac{u}{u^2+2}du + \frac{1}{4}\int \frac{1}{u^2+2}du \\= \frac{1}{8}ln|u^2+2| + \frac{1}{4} \cdot \frac{1}{\sqrt{2}}tan^{-1}(\frac{u}{\sqrt{2}})+k = \frac{1}{8}ln|4x^2-4x+3|+\frac{1}{4\sqrt{2}}tan^{-1}(\frac{2x-1}{\sqrt{2}})+k ∫ ( 2 x − 1 ) 2 + 2 x d x = ∫ u 2 + 2 2 1 ( u + 1 ) 2 1 d u = 4 1 ∫ u 2 + 2 u d u + 4 1 ∫ u 2 + 2 1 d u = 8 1 l n ∣ u 2 + 2 ∣ + 4 1 ⋅ 2 1 t a n − 1 ( 2 u ) + k = 8 1 l n ∣ 4 x 2 − 4 x + 3 ∣ + 4 2 1 t a n − 1 ( 2 2 x − 1 ) + k # Reference蘇承芳老師 - 微積分甲（一）109 學年度 - Calculus (I) Academic Year 109 