Use integration by parts to find the integral $\int\frac{\sqrt {4x^2-9}}{x^2}dx$

Question

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

I tried to solve this using integration by parts, but I come up with something that is much more difficult to solve. How can this be solved?

Answer 1

$$\int \frac{\sqrt{4x^2-9}}{x^2}dx$$ $$=-\int \sqrt{4x^2-9} \,\ d(\frac{1}{x})$$ Using by parts, $$=-\left[\frac{\sqrt{4x^2-9}}{x}-\int \frac{4}{\sqrt{4x^2-9}}dx\right]$$ $$=-\left[\frac{\sqrt{4x^2-9}}{x}-2\int \frac{d(2x)}{\sqrt{(2x)^2-3^2}}\right]$$ $$=-\left[\frac{\sqrt{4x^2-9}}{x}-2 \ln \left|2x+\sqrt{4x^2-9}\right|\right]+c$$

Answer 2

Hint. You may integrate by parts: $$ \int\frac{\sqrt {4x^2-9}}{x^2}dx=-\frac{\sqrt {4x^2-9}}{x}+8\int\frac1{\sqrt {4x^2-9}}dx $$ and you may conclude easily if you know that $$ (\text{arcsinh}\: (ax))'=\frac{a}{\sqrt {a^2x^2-1}}. $$

Answer 3

Trigonometric substitution There is another simpler method to solve the problem

Let $2x=3\sec\theta \implies dx=\frac{3}{2}\sec\theta \tan\theta \ d\theta$

$$\int \frac{\sqrt{4x^2-9}}{x^2}\ dx=\int \frac{\sqrt{9\sec^2\theta-9}}{\frac{9}{4}\sec^2\theta}\ \frac{3}{2}\sec\theta \tan\theta \ d\theta$$ taking positive value, $$=\frac{2}{3}\int \frac{3\tan\theta}{\sec\theta}\tan\theta \ d\theta$$ $$=2\int \frac{\tan^2 \theta}{\sec\theta}\ d\theta$$ $$=2\int \frac{\sec^2 \theta-1}{\sec\theta}\ d\theta$$ $$=2\int (\sec\theta-\cos\theta)\ d\theta$$

$$=2\left(\int \sec\theta\ d\theta-\int \cos\theta\ d\theta\right)$$

$$=2\left(\ln\left|\sec\theta+\tan\theta\right| -\sin\theta\right)+c$$ substituting $\sec\theta=\frac{2x}{3}$, one should get $$=2\ln|2x+\sqrt{4x^2-9}|-\frac{\sqrt{4x^2-9}}{x}+C$$

Answer 4

By substitution: $\DeclareMathOperator\ach{arg\,cosh}$

Set $ x=\dfrac32\cosh t,\enspace t\ge0$, whence $\mathrm d\mkern1mu x=\dfrac32\sinh t\,\mathrm d\mkern1mu t$. With some hyperbolic trigonometry, we get \begin{align*} \int \frac{\sqrt{4x^2-9}}{x^2}\,\mathrm d\mkern1mu x&=2\int \frac{\sinh t}{\cosh^2t}\sinh t\,\mathrm d\mkern1mu t =2\int \tanh^2t\,\mathrm d\mkern1mu t \\&=2\int\bigl(1-(1-\tanh^2t)\bigr)\,\mathrm d\mkern1mu t=2(t-\tanh t)\\ &=2\Biggl(\ach\frac{2x}3-\frac{\sinh\bigl(\ach\frac{2x}3\bigr)}{\frac{2x}3}\Biggr)\\ &=2\Biggl(\ach\frac{2x}3-\frac{3\sqrt{\frac{4x^2}9-1}}{2x}\Biggr)=2\Biggl(\ach\frac{2x}3-\frac{\sqrt{4x^2-9}}{2x}\Biggr)\\ &=2\ln\bigl(2x+\sqrt{4x^2-9}\bigr)-\frac{\sqrt{4x^2-9}}{x}+\text{constant}. \end{align*}