2
$\begingroup$

$$\int{\frac{\sqrt{x^2-3}-3\sqrt{x^2+3}}{\sqrt{x^4-9}}}dx$$

I can simplify it to: $$\int{\frac{dx}{\sqrt{x^2+3}}} - 3\int{\frac{dx}{\sqrt{x^2-3}}}$$ but I can't go from here.

$\endgroup$
1
$\begingroup$

1.$\displaystyle\int{\frac{1}{\sqrt{x^2+3}}}dx=\int\frac{1}{\sqrt3\sqrt{\frac{x^2}{\sqrt3^2}+1}}dx $

Substitute $\tan u=\frac{x}{\sqrt3}\rightarrow dx=\sqrt3\sec^2udu$, Then

$$\begin{align}\int\frac{1}{\sqrt3\sqrt{\frac{x^2}{\sqrt3^2}+1}}dx&=\int\frac{\sec^2u}{\sqrt{\tan^2u+1}}du\\&=\int \sec udu\\&=\ln(\vert \sec u+\tan u\vert)+C\\&=\ln(\vert\sec(\tan^{-1}\frac{x}{\sqrt3})+\tan(\tan^{-1}\frac{x}{\sqrt3}) \vert)+C\\&=\ln(\vert x+\sqrt{x^2+3} \vert)+C\end{align}$$

2.$\displaystyle\int{\frac{1}{\sqrt{x^2-3}}}dx=\frac{1}{\sqrt3}\cdot\int{\frac{1}{\sqrt{\frac{x^2}{\sqrt3^2}-1}}}dx$

Substitute $\sec v=\frac{x}{\sqrt3}\rightarrow dx=\sqrt3\tan v\sec vdv$, Then

$$\begin{align} \frac{1}{\sqrt3}\cdot\int{\frac{1}{\sqrt{\frac{x^2}{\sqrt3^2}-1}}}dx&=\int\frac{\tan v\sec v}{\tan v}dv\\&=\int\sec v dv\\&=\ln(\vert\sec v+\tan v\vert)+C\\&=\ln(\vert\sec(\sec^{-1} \frac{x}{\sqrt3})+\tan(\sec^{-1}\frac{x}{\sqrt3})\vert)+C\\ &=\ln(\vert x+\sqrt{x^2-3}\vert)+C \end{align}$$

$\endgroup$
0
$\begingroup$

Hint : $$\int{\frac{dx}{\sqrt{x^2+3}}} - 3\int{\frac{dx}{\sqrt{x^2-3}}}\\=\int{\frac{dx}{\sqrt{3}\sqrt{(\frac{x}{\sqrt3})^2+1}}} - 3\int{\frac{dx}{\sqrt{3}\sqrt{(\frac{x}{\sqrt3})^2-1}}}$$

$\endgroup$
  • $\begingroup$ You can recognise formulas which are often in tables of integrals $\endgroup$ – Jennifer Feb 7 '17 at 12:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.