Question

If $\int (x^6+x^4+x^2)\sqrt{2x^4+3x^2+6}dx=\frac{(\alpha x^6+\beta x^4+\gamma x^2{)}^{3/2}}{18}+C$ where C is constant then find the value of $(\beta +\gamma -\alpha ).$

Answer 1

$\int (x^6+x^4+x^2)\sqrt{2x^4+3x^2+6}dx=\int (x^5+x^3+x)\sqrt{2x^6+3x^4+6x^2}dx$

Let $2x^6+3x^4+6x^2=t$

$\therefore dt=12(x^5+x^3+x)$

$⟹\int (x^6+x^4+x^2)\sqrt{2x^4+3x^2+6}dx=\int \frac{1}{12}\sqrt{t}dt$

$=\frac{1}{12}\times \frac{2}{3}{t}^{\frac{3}{2}}+c$

$=\frac{1}{18}(2x^6+3x^4+6x^2{)}^{\frac{3}{2}}+c$

$⟹\beta +\gamma -\alpha =7$