intergrals

**Count Hesabu** · April 15th 2008, 12:34 AM

integral of (dx)/(x(x^2+5))

**angel.white** · April 15th 2008, 01:23 AM

Originally Posted by Count Hesabu

integral of (dx)/(x(x^2+5))

1. Use a partial fraction decomposition

$=\frac 15 \int\left( \frac 1x - \frac x{x^2+5} \right) ~dx$

2. Split up the integral

$=\frac 15 \int \frac 1x ~dx - \frac 15 \int \frac x{x^2+5} ~dx$

3. Make a smart substitution

-----------------------------
Substitute:
$a = x^2 +5$

$\frac 12 ~da = x ~dx$
-----------------------------

$=\frac 15 \int \frac 1x ~dx - \frac 1{10} \int \frac 1{a} ~da$

4. Integrate

$=\frac 15 ln|x| - \frac 1{10} ln|a| +C$

5. Anti-substitute

$=\frac 15 ln|x| - \frac 1{10} ln|x^2+5| +C$

6. Square and square root the x on the left

$=\frac 15 ln|(x^2)^{1/2}| - \frac 1{10} ln|x^2+5| +C$

7. Pull the exponent into the coefficient

$=\frac 1{10} ln|x^2| - \frac 1{10} ln|x^2+5| +C$

8. Combine logs

$=\frac 1{10} ln\left|\frac {x^2}{x^2+5}\right| +C$

**Count Hesabu** · April 15th 2008, 03:05 AM

thanks it helped

**polymerase** · April 15th 2008, 06:07 AM

Originally Posted by Count Hesabu

integral of (dx)/(x(x^2+5))

$\int \frac{1}{x(x^2+5)}\;dx=\int \frac{x^2-(x^2+5)}{x(x^2+5)}\;dx=\frac{1}{5}\int \frac{1}{x}\;dx-\frac{1}{5}\int \frac{x}{x^2+5}\;dx$ $=\frac{1}{5}\ln|x|-\frac{1}{10}\ln|x^2+5|$

Thus, $\int \frac{1}{x(x^2+5)}\;dx=\frac{1}{10}\ln\left(\frac{ x^2}{x^2+5}\right)+C$

Note: no need for absolute bars anymore because $x^2$ and $x^2+5$ will never be negative

**angel.white** · April 15th 2008, 08:22 AM

Originally Posted by polymerase

Note: no need for absolute bars anymore because $x^2$ and $x^2+5$ will never be negative

Good point.

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