You cannot expand it further without the use of Maclaurin series’, even then you will get an infinite series that only converges if certain conditions are met, not a finite expression. If it were a ...

Note that by continuity \lim_{x\to 0} \sqrt{1+x+x^2}=\sqrt{1+0+0^2}=1 \lim_{x\to 1} \sqrt x=\sqrt 1=1 and since \lim_{x\to 1} \sqrt x=\lim_{x\to 0} \sqrt {1+x}=1 then \lim_{x\to 0} \frac{\sqrt{1+x+x^2}}{ \sqrt {1+x}}=\sqrt{1+\frac{x^2}{1+x}}=1 ...

\begin{align} \int\sqrt{x^4+x^2}\,dx&=\int\sqrt{x^2(x^2+1)}\,dx\\ &=\int x\sqrt{x^2+1}\,dx\\ &=\int \sqrt{x^2+1}\,xdx\\ \end{align} Let u=x^2+1, then du=2x\,dx or x\,dx=\frac{1}{2}du. Hence ...

Don't forget the absolute value. \sqrt{2}\int_0^9|2x-6|dx=\sqrt{2}\left(\int_0^3(6-2x)dx+\int_3^9(2x-6)dx\right) which is equal to \sqrt{2}\left(6x-x^2|_0^3+x^2-6x|_3^9\right)=45\sqrt{2}

I hope you do not mind if I prefer to start from scratch. We have \int (2x+1)\sqrt{x^2+x}\,dx = C+\frac{2}{3}(x^2+x)^{3/2} \tag{1} and the problem boils down to computing \int\sqrt{x^2+x}\,dx. ...

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