User talk:Gumby701/calc

$$\int x^5 \sqrt{{x^2}-1} dx = \int (x^2)^2 \sqrt{{x^2}-1}\,xdx$$

Using rule number 60:

$$\int u^n \sqrt = \frac{2}{b(2n+3)} \left \lbrack u^n(a+bu)^{3/2} -na \int u^{n-1} \sqrt{a+bu} \, du \right \rbrack$$

where...

$$\,\!u = x^2$$

$$\,\!du = 2xdx$$

$$\,\!xdx = \frac{du}{2}$$

$$\,\!n = 2$$

$$\,\!a = -1$$

$$\,\!b = 1$$

$$ \begin{align} \int (x^2)^2 \sqrt{{x^2}-1}\,xdx & = \frac{1}{2} * \frac{2}{1(2*2+3)} \left \lbrack {x^4}(x^2-1)^{3/2} -(2*{^-}1) \int x^{2({2-1})} \sqrt{{x^2}-1}\,dx\right \rbrack \\ & = \frac{1}{7} \left \lbrack {x^4}(x^2-1)^{3/2} + 2 \int x^{2} \sqrt{{x^2}-1}\,dx\right \rbrack \\ & = \frac{{x^4}(x^2-1)^{3/2}}{7} + \frac{2}{7} \int x^{2} \sqrt{{x^2}-1}\,dx \\ \end{align} $$

Using rule number 40:

$$\int u^2 \sqrt{u^2-a^2}\,du = \frac{u}{8} (2u^2 - a^2) \sqrt{u^2 - a^2} - \frac{a^4}{8}\ln \Big|u + \sqrt{u^2 + a^2} \, \Big| + C $$

$$\,\!u^2 = x^2$$

$$\,\!u = x$$

$$\,\!du = dx$$

$$\,\!a^2 = 1$$

$$\,\!a = 1$$

$$ \begin{align} \frac{{x^4}(x^2-1)^{3/2}}{7} + \frac{2}{7} \int x^{2} \sqrt{{x^2}-1}\,dx & = \frac{{x^4}(x^2-1)^{3/2}}{7} + \frac{2}{7} \left \lbrack \frac{x}{8} (2x^2 - 1) \sqrt{x^2 - 1} - \frac{1}{8}\ln \Big|x + \sqrt{x^2 + 1} \, \Big| + C \right \rbrack \\ & = \frac{{x^4}(x^2-1)^{3/2}}{7} + \frac{x (2x^2 - 1) \sqrt{x^2 - 1}}{28} - \frac{\ln \Big|x + \sqrt{x^2 + 1} \, \Big|}{28} + C \\ & = \frac{{x^4}(x^2-1)^{3/2}}{7} + \frac{x (x^2-1)^{3/2}}{28} - \frac{\ln \Big|x + \sqrt{x^2 + 1} \, \Big|}{28} + C \\ & = \frac{4{x^4}(x^2-1)^{3/2}}{28} + \frac{x (x^2-1)^{3/2}}{28} - \frac{\ln \Big|x + \sqrt{x^2 + 1} \, \Big|}{28} + C \\ & = \frac{({x^4}+ x)(x^2-1)^{3/2}}{7} - \frac{\ln \Big|x + \sqrt{x^2 + 1} \, \Big|}{28} + C \\ & = \frac{x({x^3}+ 1)(x^2-1)^{3/2}}{7} - \frac{\ln \Big|x + \sqrt{x^2 + 1} \, \Big|}{28} + C \\ \end{align} $$