Priciple
With goes down to those hard problems, I want to let you know a priciple that we use in general math problem:
We always split hard problem to multiple easy problems or step by step simplify hard problem.
In computer science, we named it Divide & Conquer (D&C algorithms)
Simple one
ā« 1 a 2 + x 2 d x = 1 a 2 ā« 1 1 + ( x a ) 2 d x //divideĀ a 2 inĀ denominatorĀ whileĀ keepĀ equationĀ balance = 1 a ā« 1 1 + ( x a ) 2 d ( x a ) = 1 a ā
arctan ā” x a + C // ( arctan ā” x ) ā² = 1 1 + x 2 \begin{align*}
&\int \frac{1}{a^2 + x^2} dx
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=& \frac{1}{a^2} \int \frac{1}{1 + (\frac{x}{a})^2} dx &\text{//divide } a^2 \text{in denominator while keep equation balance}
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=& \frac{1}{a} \int \frac{1}{1 + (\frac{x}{a})^2} d(\frac{x}{a})
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=& \frac{1}{a} \cdot \arctan{\frac{x}{a}} + C &\text{//}(\arctan{x})^\prime = \frac{1}{1 + x^2}
\end{align*} = = = ā ā« a 2 + x 2 1 ā d x a 2 1 ā ā« 1 + ( a x ā ) 2 1 ā d x a 1 ā ā« 1 + ( a x ā ) 2 1 ā d ( a x ā ) a 1 ā ā
arctan a x ā + C ā //divideĀ a 2 inĀ denominatorĀ whileĀ keepĀ equationĀ balance // ( arctan x ) ā² = 1 + x 2 1 ā ā ā« 1 a 2 ā x 2 d x , ( a > 0 ) = 1 a ā« 1 1 ā ( x a ) 2 d x = 1 ā« 1 1 ā ( x a ) 2 d ( x a ) //divideĀ a Ā inĀ xĀ ofĀ dxĀ whileĀ keepĀ equationĀ balance = arcsin ā” ( x a ) + C \begin{align*}
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&\int \frac{1}{\sqrt{a^2 - x^2}} dx ,(a > 0)
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=& \frac{1}{a} \int \frac{1}{\sqrt{1-(\frac{x}{a})^2}} dx
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=& 1 \int \frac{1}{\sqrt{1-(\frac{x}{a})^2}} d(\frac{x}{a}) &\text{//divide } a \text{ in x of dx while keep equation balance}
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=& \arcsin(\frac{x}{a}) + C
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\end{align*} = = = ā ā« a 2 ā x 2 ā 1 ā d x , ( a > 0 ) a 1 ā ā« 1 ā ( a x ā ) 2 ā 1 ā d x 1 ā« 1 ā ( a x ā ) 2 ā 1 ā d ( a x ā ) arcsin ( a x ā ) + C ā //divideĀ a Ā inĀ xĀ ofĀ dxĀ whileĀ keepĀ equationĀ balance ā Let's try to analyze this equation:
1 a 2 ā« 1 1 + ( x a ) 2 d x = a a 2 ā« 1 1 + ( x a ) 2 d ( x a ) \frac{1}{a^2} \int \frac{1}{1 + (\frac{x}{a})^2} dx = \frac{a}{a^2} \int \frac{1}{1 + (\frac{x}{a})^2} d(\frac{x}{a}) a 2 1 ā ā« 1 + ( a x ā ) 2 1 ā d x = a 2 a ā ā« 1 + ( a x ā ) 2 1 ā d ( a x ā ) We knew that x x x d x dx d x 1 a 2 \frac{1}{a^2} a 2 1 ā x x x x a \frac{x}{a} a x ā 1 a 2 \frac{1}{a^2} a 2 1 ā a a a
As for ā« 1 1 + ( x a ) 2 \int \frac{1}{1 + (\frac{x}{a})^2} ā« 1 + ( a x ā ) 2 1 ā d x dx d x
You can see ā« ( . . . ) d \int (...) d ā« ( ... ) d
ā« 1 x 2 ā a 2 d x , ( a ā 0 ) = ā« 1 ( x + a ) ( x ā a ) d x = 1 2 a ā« ( 1 x ā a ā 1 x + a ) d x // 1 x ā a ā 1 x + a = x + a ā x + a ( x ā a ) ( x + a ) = 2 a ( x ā a ) ( x + a ) // Ā BecauseĀ 2 a ā 1 ,Ā divideĀ 2a,Ā thatĀ is,Ā times 1 2 a inĀ theĀ beginning = 1 2 a ( ā« 1 x ā a d x ā ā« 1 x + a d x ) = 1 2 a ( ā« 1 x ā a d ( x ā a ) ā ā« 1 x + a d ( x ā a ) ) //donātĀ forgetĀ theĀ basicĀ templeĀ forĀ solvingĀ integralĀ eqution = 1 2 a ln ā” ā£ x ā a x + a ā£ + C \begin{align*}
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&\int \frac{1}{x^2 - a^2} dx ,(a \neq 0)
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=& \int \frac{1}{(x+a)(x-a)} dx
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=& \frac{1}{2a} \int (\frac{1}{x-a} - \frac{1}{x+a}) dx
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\text{//}& \frac{1}{x-a} - \frac{1}{x+a} = \frac{x+a-x+a}{(x-a)(x+a)} = \frac{2a}{(x-a)(x+a)}
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\text{//}&\text{ Because }2a \neq 1 \text{, divide 2a, that is, times} \frac{1}{2a} \text{in the beginning}
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=& \frac{1}{2a}(\int \frac{1}{x-a} dx - \int \frac{1}{x+a} dx)
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=& \frac{1}{2a}(\int \frac{1}{x-a} d(x-a) - \int \frac{1}{x+a} d(x-a)) &\text{//don't forget the basic temple for solving integral eqution}
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=& \frac{1}{2a} \ln{|\frac{x-a}{x+a}|} + C
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\end{align*} = = // // = = = ā ā« x 2 ā a 2 1 ā d x , ( a ī = 0 ) ā« ( x + a ) ( x ā a ) 1 ā d x 2 a 1 ā ā« ( x ā a 1 ā ā x + a 1 ā ) d x x ā a 1 ā ā x + a 1 ā = ( x ā a ) ( x + a ) x + a ā x + a ā = ( x ā a ) ( x + a ) 2 a ā Ā BecauseĀ 2 a ī = 1 ,Ā divideĀ 2a,Ā thatĀ is,Ā times 2 a 1 ā inĀ theĀ beginning 2 a 1 ā ( ā« x ā a 1 ā d x ā ā« x + a 1 ā d x ) 2 a 1 ā ( ā« x ā a 1 ā d ( x ā a ) ā ā« x + a 1 ā d ( x ā a )) 2 a 1 ā ln ā£ x + a x ā a ā ā£ + C ā //donātĀ forgetĀ theĀ basicĀ templeĀ forĀ solvingĀ integralĀ eqution ā Complex one
ā« 1 1 ā x d x makeĀ x = t Ā ,Ā thenĀ x = t 2 = ā« 1 1 ā t d ( t 2 ) = ā« 2 t 1 ā t d ( t ) = 2 ā« t 1 ā t d ( t ) = ā 2 ā« t t ā 1 d ( t ) = ā 2 ā« t ā 1 + 1 t ā 1 d ( t ) = ā 2 ā« t ā 1 t ā 1 d t ā 2 ā« 1 t ā 1 d t = ā 2 ā« 1 d t ā 2 ā« 1 t ā 1 d t = ā 2 t ā 2 ln ā” ā£ t ā 1 ā£ + C = ā 2 x ā 2 ln ā” ā£ x ā 1 ā£ + C \begin{align*}
&\int \frac{1}{1 - \sqrt{x}} dx
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& \text{make } \sqrt{x}=t \text{ , then } x=t^2
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=& \int \frac{1}{1-t} d(t^2)
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=& \int \frac{2t}{1-t} d(t)
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=& 2 \int \frac{t}{1-t} d(t)
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=& -2 \int \frac{t}{t-1} d(t)
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=& -2 \int \frac{t-1+1}{t-1} d(t)
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=& -2 \int \frac{t-1}{t-1} dt -2 \int \frac{1}{t-1} dt
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=& -2 \int 1 dt -2 \int \frac{1}{t-1} dt
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=& -2t - 2 \ln{|t-1|} + C
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=& -2\sqrt{x} - 2 \ln{|\sqrt{x}-1|} + C
\end{align*} = = = = = = = = = ā ā« 1 ā x ā 1 ā d x makeĀ x ā = t Ā ,Ā thenĀ x = t 2 ā« 1 ā t 1 ā d ( t 2 ) ā« 1 ā t 2 t ā d ( t ) 2 ā« 1 ā t t ā d ( t ) ā 2 ā« t ā 1 t ā d ( t ) ā 2 ā« t ā 1 t ā 1 + 1 ā d ( t ) ā 2 ā« t ā 1 t ā 1 ā d t ā 2 ā« t ā 1 1 ā d t ā 2 ā« 1 d t ā 2 ā« t ā 1 1 ā d t ā 2 t ā 2 ln ā£ t ā 1ā£ + C ā 2 x ā ā 2 ln ā£ x ā ā 1ā£ + C ā Conclusion
Best way to deal with \sqrt{} ā
Molecular is a constant is more convenient for us to solve equation, especially when it is 1 1 1 ā« \int ā«