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How I can evaluate the indefinite integral? :
$$\int\frac{1}{x^2-6x}\,\mathrm dx$$
Do I need to bring it to this format? : $\displaystyle \int\frac{1}{x^2-a^2}\,\mathrm dx\;$?
Thanks!

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5  
The magic words are "partial fractions". –  David Mitra Apr 30 '13 at 17:19
2  
Hint: $(x-3)^2=x^2-6x+9$. –  Thomas Andrews Apr 30 '13 at 17:22
    
You can bring it in that format, or you do partial fraction decomposition. You answer is going to be consisting of two natural log terms with the denominators of your fractions inside it.. Now give it a try... –  imranfat Apr 30 '13 at 17:22
    
ditto on partial fractions –  Stefan Smith May 1 '13 at 2:07

6 Answers 6

up vote 8 down vote accepted

$$\int\frac{1}{x^2-6x}\,\mathrm dx = \int\frac{1}{x(x-6)}\,\mathrm dx$$

Now use partial fraction decomposition to obtain an integral of the form $$\int\frac{1}{x(x-6)}\,\mathrm dx =\int \left(\frac{A}{x} + \frac{B}{x-6}\right)\,\mathrm dx$$

Now all you need to do is to determine the (constant) values of $A$ and of $B$ so that $$ \left(\frac{A}{x} + \frac{B}{x-6}\right) = \frac{1}{x(x-6)}.$$

$$\left(\frac{A}{x} + \frac{B}{x-6}\right) = \frac{A(x-6) + B(x)}{x(x - 6)} \iff A(x - 6) + B(x) = 1\iff (A + B)x - 6A = 1 $$

$$\iff A+B = 0 \;\;\text{and}\;\; -6A = 1 \iff \color{blue}{\bf A = -\frac 16}, \;\;\text{and}\;\; B = - A \iff \color{red}{\bf B = \frac 16}$$

This gives us, then, Now use $$\int\frac{1}{x(x-6)}\,\mathrm dx =\int \left(\frac{A}{x} + \frac{B}{x-6}\right)\,\mathrm dx = \int \left(\color{blue}{\bf -}\frac{\color{blue}{\bf 1}}{\color{blue}{\bf 6}x} + \frac{\color{red}{\bf 1}}{\color{red}{\bf 6}(x-6)}\right)\,\mathrm dx$$ $$ = -\frac 16 \int \left(\dfrac 1x \right) \,\mathrm dx + \frac 16 \int \left(\frac 1{x-6}\right)\,\mathrm dx$$

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Hi Angel Amy. :-) –  B. S. May 1 '13 at 14:54
    
Hello, my dear friend! ;-) –  amWhy May 1 '13 at 14:54
1  
Yes, exactly...they mean the same thing. I'd suggest simply "where a and b are both constants..." –  amWhy May 1 '13 at 15:02
    
Thanks a lot. I am preparing my paper right now. –  B. S. May 1 '13 at 15:02
    
Thanks for your kind and supportinng words Amy. Yes I have an profile at FB but don't have enough time to get there. :-( –  B. S. May 1 '13 at 15:06

The obvious method is partial fractions, but if you desperately want it in the form $\displaystyle \int \dfrac{1}{x^2-a^2}\, dx$ then complete the square on the denominator and substitute $u=x-3$; you get $$\int \dfrac{1}{x^2-6x}\, dx = \int \dfrac{1}{u^2-9}\, du$$

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Use partial fraction decomposition to obtain a sum of easily integrable terms:

$$\frac{1}{x^2-6x}=\frac{1}{x(x-6)}=\frac{A}{x}+\frac{B}{x-6}$$

$$\text{multiply by }x \implies \frac{1}{x-6}=A+\frac{Bx}{x-6},\quad \text{set }x=0 \implies \color{blue}{A=-\frac{1}{6}}$$

$$\text{multiply by }x-6 \implies \frac{1}{x}=\frac{A(x-6)}{x}+B,\quad \text{set }x=6 \implies \color{red}{B=\frac{1}{6}}$$

Hence

$$\int\frac{1}{x^2-6x}dx=\int\left(\color{blue}-\frac{\color{blue}1}{\color{blue}6x}+\frac{\color{red}1}{\color{red}6(x-6)}\right)dx$$

Now, use the fact that

$$\int\frac{1}{x-a}dx=\ln|x-a|$$

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$\int\frac{1}{x^2-a^2}\,\mathrm dx=\frac{1}{2a}\int\frac{1}{x-a}-\frac{1}{x+a}dx=\frac{1}{2a}(ln(x-a) -ln(x+a))$

and about $\int \frac{1}{x^2-6x}=\frac{-1}{6}\int \frac{1}{x}-\frac{1}{x-6}dx=\frac{-1}{6}(lnx-ln (x-6))$

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$$\frac{1}{x^2-6x}=\frac{1}{x(x-6)}=\frac{A}{x}+\frac{B}{x-6}$$

$$1=A(x-6)+Bx$$ $$(A+B)x-6A=1$$ $$-6A=1,A+B=0 $$ $$A=-1/6,B=1/6$$

$$\int\frac{1}{x^2-6x}dx=\frac{1}{6}\int\left(\frac{1}{x-6}-\frac{1}{x}\right)dx=$$ $$=\frac{1}{6}(\ln|x-6|-\ln|x|)+C$$

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$+1$ for missed FB member Adi –  B. S. May 1 '13 at 15:00

I guess I'm late)

Denote $x-3=t$ $$ \int \frac{dx}{x^2-6x}=\int \frac{dx}{x^2-6x +9-9}=\int \frac{dt}{(t-3)(t+3)}=\frac{1}{6} \int \frac{dt}{t-3} -\frac{1}{6} \int \frac{dt}{t+3}=\frac{1}{6}\log \Bigg|\frac{x-6}{x} \bigg| $$

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