SOLUTION 11: $\ \$ To integrate $\displaystyle{ \int { x \over \sqrt{x^4-16} } \ dx = \int { x \over \sqrt{(x^2)^2-16} } \ dx }$ begin with the ordinary u-substitution $$u = x ^{2}$$ so that $$du = 2x \ dx$$ Substitute into the original problem, replacing all forms of x, getting $$\displaystyle { \int \frac{x}{\sqrt{x^{4}-16}} \ dx = \int \frac{1/2}{\sqrt{u^{2}-16}} \ du }$$ Now use the trig substitution $$u = 4 \sec \theta$$ so that $$du = 4 \sec \theta \tan \theta \ d \theta$$ Substitute into the original problem, replacing all forms of u, getting $$= \displaystyle { \int \frac{1/2}{\sqrt{(4 \sec \theta )^{2}-16}} \ du }$$ $$= \displaystyle { \int \frac{1/2}{\sqrt{16 \sec^{2} \theta - 16}} \cdot 4 \sec \theta \tan \theta \ d \theta }$$ $$= \displaystyle { \int \frac{1/2}{\sqrt{16 ( \sec^{2} \theta - 1)}} \cdot 4 \sec \theta \tan \theta \ d \theta }$$ $$= \displaystyle { 2 \int \frac{\sec \theta \tan \theta}{\sqrt{16 \tan^{2} \theta}} \ d \theta }$$ $$= \displaystyle { 2 \int \frac{\sec \theta \tan \theta}{4 \tan \theta} \ d \theta }$$ $$= \displaystyle { \frac{1}{2} \int \sec \theta \ d \theta }$$ $$= \displaystyle { \frac{1}{2} (\ln \Big|\sec \theta + \tan \theta \Big| + C) }$$ $\Big($ We need to write our final answer in terms of $x$.

Since $u = 4 \sec \theta$ it follows that $$\sec \theta = \displaystyle{ u \over 4 } = \displaystyle{ hypotenuse \over adjacent }$$ and from the Pythagorean Theorem that $$\displaystyle (adjacent)^2 + (opposite)^2 = (hypotenuse)^2 \ \ \longrightarrow$$ $$(4)^2 + (opposite)^2 = (u)^2 \ \ \longrightarrow \ \ \ opposite = \sqrt{u^2-16} \ \ \longrightarrow$$ $$\tan \theta = \displaystyle{ opposite \over adjacent }= \displaystyle{ \sqrt{u^2-16} \over 4 } . \Big)$$ $$= \displaystyle { \frac{1}{2} (\ln \Big|{ u \over 4 } + { \sqrt{u^2-16} \over 4 }\Big| + C) }$$ (Now use the fact that $u=x^2$.) $$= \displaystyle { \frac{1}{2} (\ln \Big|\frac{x^{2}}{4} + \frac{\sqrt{x^{4}-16}}{4} \Big| + C) }$$