1. ## Intregration with Table

How do you intergrate dx/x^2(4x^2+9)?

thanks

2. Ignore the constants,

Do you know how to integrate,
$\int \frac{1}{x^2\sqrt{1+x^2}}dx$

Just use the substitution $t=\tan^{-1} x$.

3. $\int\frac{1}{x^{2}(4x^{2}+9)}dx$

Expand into partial fractions:

$\frac{1}{9}\int\frac{1}{x^{2}}dx-\frac{4}{9}\int\frac{1}{4x^{2}+9}dx$

The first part is easy. You can use a substitution for the second one.

Try using $u=tan^{-1}(\frac{2x}{3}), \;\ du=\frac{6}{4x^{2}+9}dx$

4. $\displaystyle \int\frac{dx}{x^2+a^2}=\frac{1}{a}\arctan\frac{x}{ a}+C$

$\displaystyle \int\frac{dx}{4x^2+9}=\frac{1}{4}\int\frac{dx}{x^2 +\left(\frac{3}{2}\right)^2}=\frac{1}{4}\cdot\frac {2}{3}\arctan\frac{2x}{3}+C$

$\int\frac{dx}{x^2(4x^2+9)}$
Let $2x \:=\:3\tan\theta\quad\Rightarrow\quad x \:=\:\frac{3}{2}\tan\theta\quad\Rightarrow\quad dx \:=\;\frac{3}{2}\sec^2\!\theta\,d\theta$

. . $4x^2 + 9 \; = \; 9\tan^2\!\theta + 9 \; = \; 9(\tan^2\!\theta + 1) \; = \;9\sec^2\!\theta$

Substitute: . $\int\frac{\frac{3}{2}\sec^2\!\theta\,d\theta}{\fra c{9}{4}\tan^2\!\theta\cdot9\sec^2\!\theta} \;=\;\frac{2}{27}\int\frac{d\theta}{\tan^2\theta} \;= \; \frac{2}{27}\int\cot^2\theta\,d\theta$

. . $= \;\frac{2}{27}\int\left(\csc^2\theta - 1\right)\,d\theta \;=\;\frac{2}{27}\left(-\cot\theta - \theta\right) + C\;=\;-\frac{2}{27}(\cot\theta + \theta) + C$

Back-substitute: . $-\frac{2}{27}\left[\frac{3}{2x} \,+ \,\arctan\!\left(\frac{2x}{3}\right)\right] + C$