Trigonometric substitution hint

Question

I don't know how to get the following problem started and would appreciate a hint:

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

Answer 1

write the integral as :

$$I=\int \frac{dx}{x^3 \sqrt{9+\dfrac{4} {x^2}}}$$

Now you can subsitute t = $9 + \dfrac{4} {x^2}$

$$ \begin{align} dt &= -\dfrac{4}{x^3}dx \\ \dfrac{dt}{-8} &= \dfrac{dx}{x^3} \\ I&=\int \frac{dt}{-8 \sqrt{t}} \\ \end{align} $$ This can be easily integrated !

In problems like these, sometimes you can get a very neat answer by taking x^n from the root.( I am unable to express it better and if you can write it better, please edit the answer.)

Answer 2

$\int \dfrac{1}{3x^{2}\sqrt{x^{2}+\dfrac{4}{9}}}$

let $x=\dfrac{2}{3}tan(u)$ so $dx=\dfrac{2}{3}(\sec^2 u)du$ $$I=\int \dfrac{\frac{2}{3}(\sec^{2}u)}{3\frac{4}{9} \tan^2u\sqrt{\frac{4}{9} \tan^2u + \frac{4}{9}}}du$$

$$=\int \frac{\sec^2 u }{\frac{4}{3} \tan^2u \sqrt{\sec^2u}}$$ $$=\frac{3}{4} \csc u\cot u du = -\frac{3}{4} \csc u + c$$

since $x= \frac{2}{3}\tan u$ it follows (draw a triangle) that $\csc(u)=\frac{\sqrt{9x^2+4}}{3x}$ and so the integral reduces to $$-\frac{\sqrt{9x^2+4}}{4x} +C$$

Answer 3

Jagy's substitution is the best here

substitute $x=\frac 23 \sinh(t)$ and $ dx=\frac 23 \cosh(t)dt$ $$I=\int \frac{1}{x^2 \sqrt{9x^2+4}}dx$$ $$\int \frac{\frac 23 \cosh(t)}{\frac 49\sinh^2(t) \sqrt{4\sinh^2(t)+4}}dt$$ $$\int \frac{\cosh(t)}{\frac 23\sinh^2(t) 2\cosh(t)}dt$$ $$I=\frac 34\int \frac{dt}{\sinh^2(t) }=-\frac 34\coth(t)$$ Substitute back now...