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Hi I have the following integral: $$\int \frac{2x}{x^2+6x+3}\, dx$$

I made some changes like: $$\int \dfrac{2x+6-6}{x^2+6x+3}\, dx$$

then I have: $$\int \dfrac{2x+6}{x^2+6x+3}\, dx -\int\dfrac{6}{x^2+6x+3}\, dx$$

and thus: $$\ln(x^2+6x+3)-\int\dfrac{6}{x^2+6x+3}\, dx$$

Ok, I have decomposed $$\frac{2x}{x^2+6x+3} $$ in: $$ \frac{3+\sqrt6}{\sqrt6(x+\sqrt 6+3)} + \frac{3-\sqrt6}{\sqrt6 (-x+\sqrt6-3)}$$

How can I integrate this expressions?

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Hello, please use mathjax to improve readability. Concerning your question, you can put it in the form $\frac{\alpha}{x^2 +1}$ which you will recognize as arctangente or $\frac{\alpha}{x^2 - 1}$ that you can decompose in fractions as David suggested. –  Matt B. Jul 30 at 14:51
    
Hint in response to your most recent edit: natural log with substitution. –  anorton Jul 30 at 16:11

4 Answers 4

A start: Note that $x^2+6x+3=0$ has the roots $\alpha=-3+\sqrt{6}$ and $\beta=-3-\sqrt{6}$. Thus $x^2+6x+3=(x-\alpha)(x-\beta)$.

Express $\frac{6}{(x-\alpha)(x-\beta)}$ as $\frac{A}{x-\alpha}+\frac{B}{(x-\beta)}$ (partial fractions).

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$$\frac{2x}{x^2+6x+3}=\frac{A}{x-(-3-\sqrt{6})}+\frac{B}{x-(\sqrt{6}-3)}$$

Find $A$ and $B$ and then:

$$\int \frac{2x}{x^2+6x+3} dx=\int \frac{A}{x-(-3-\sqrt{6})} dx+\int \frac{B}{x-(\sqrt{6}-3)} dx$$

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HINT:

As $\displaystyle x^2+6x+3=(x+3)^2-(\sqrt6)^2,$

using Trigonometric substitution, set $x+3=\sqrt6\sec\theta$

or use $\#1$ of this

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Another idea (just reducing it to another form):

Let $$I=6\int \frac{1}{x^2+6x+3} dx=6\int \frac{1}{(x+3)^2-6} dx=\int \frac{1}{(\frac{1}{\sqrt{6}}(x+3))^2-1} dx$$.

Now let $$\frac{1}{\sqrt{6}}(x+3)=\cosh a$$, hence using $$\cosh^2 a - 1 = \sinh ^2 a$$ and $$\frac{1}{\sqrt{6}} = \sinh a \frac{da}{dx}\Leftrightarrow dx = da \sinh a \sqrt{6}$$ we get $$I=\int \frac{1}{\sinh ^2 a} da \sinh a\sqrt{6} = \sqrt{6} \int \frac{1}{\sinh a} da$$.

EDIT: Can someone please show me how to write bigger LaTeX? EDIT2: Neat!

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1  
Put stuff between double dollar signs: $\$\$$ stuff $\$\$$. This will put the equation on a line by itself. If you want it big "inline", use "\displaystyle": $\$$\displaystyle{this is big stuff}$\$ $. Be careful, it may be too big. –  David Mitra Jul 30 at 15:04
    
Thanks! @DavidMitra –  Eulerian Adventurer Jul 30 at 15:38

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