# Which integration techniques I should use for $\int{\frac{\sqrt{x^2-3}-3\sqrt{x^2+3}}{\sqrt{x^4-9}}}dx$

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

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.

• – lab bhattacharjee Feb 7 '17 at 11:28
• substitute $\tan u=\frac{x}{\sqrt3}$ and $\sec u=\frac{x}{\sqrt3}$ respectively. – MoNtiDeaD MoonDogs Feb 7 '17 at 11:28

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}

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

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