1. intergrals

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

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

$\frac 12 ~da = x ~dx$
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$=\frac 15 \int \frac 1x ~dx - \frac 1{10} \int \frac 1{a} ~da$

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

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

3. thanks it helped

4. 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

5. Originally Posted by polymerase
Note: no need for absolute bars anymore because $x^2$ and $x^2+5$ will never be negative
Good point.