<s>[INST] Find integral of $\int\frac{x^2+3}{\sqrt{x^2+6}}dx$ from Cambridge IGCSE Additional Mathematics $$\int\frac{x^2+3}{\sqrt{x^2+6}}dx$$
I tried to $$\int\frac{x^2+6-3}{\sqrt{x^2+6}}dx=\int\frac{x^2+6}{\sqrt{x^2+6}}dx-\int\frac{3}{\sqrt{x^2+6}}dx= \int\sqrt{x^2+6}dx - 3\int\frac{1}{\sqrt{x^2+6}}dx$$
Then got stuck
In exam-mate website the answer is $$\frac{1}{2}x\sqrt{x^{2}+6}$$
Thanks for replies in advance
 [/INST] You can enforce an Euler substitution,
$$t=\frac{\sqrt{x^2+6}-\sqrt6}x \implies x=\frac{2\sqrt6\,t}{1-t^2} \implies dx = 2\sqrt6\frac{1+t^2}{(1-t^2)^2} \, dt$$
to transform the starting integral to
$$\int \frac{x^2+3}{\sqrt{x^2+6}}\,dx = 2\sqrt6 \int \frac{\frac{24t^2}{(1-t^2)^2}+3}{\frac{2\sqrt6\,t^2}{1-t^2}+\sqrt6} \, \frac{1+t^2}{(1-t^2)^2}\,dt = 6 \int \frac{1+6t^2+t^4}{(1-t^2)^3}\,dt$$
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