SOLUTION 15: $\ \$ Integrate $\displaystyle{ \int { \sqrt{ {x-1} \over x } } \ dx = \int { \frac{ \sqrt{x-1} }{ \sqrt{ x} } } \ dx }$. Begin with the power substitution $$x = u^{2}$$ so that $$dx = 2u \ du$$ Substitute into the original problem, replacing all forms of $x$, getting $$\displaystyle { \int { \frac{ \sqrt{x-1} }{ \sqrt{ x} } } \ dx = \int \frac{ \sqrt{u^2-1} }{ \sqrt{u^2} } \cdot 2u \ du }$$ $$= \displaystyle { { 2 \int \frac{ \sqrt{u^2-1} }{ u} } \cdot u \ du }$$ $$= \displaystyle { { 2 \int { \sqrt{u^2-1} } } \ du }$$ Now use the trig substitution $$u = \sec \theta$$ so that $$du = \sec \theta \tan \theta \ d \theta$$ Substitute into the original problem, replacing all forms of $u$, getting $$\displaystyle { { 2 \int { \sqrt{u^2-1} } } \ du } = \displaystyle { 2 \int \sqrt{\sec^2 \theta -1} \sec \theta \tan \theta \ d \theta }$$ $$= \displaystyle { 2 \int \sqrt{\tan^2 \theta } \sec \theta \tan \theta \ d \theta }$$ $$= \displaystyle { 2 \int \tan \theta \sec \theta \tan \theta \ d \theta }$$ Now use integration by parts. Let $u = \tan \theta$ and $dv = \sec \theta \tan \theta \ d \theta$ , so that $du = \sec^2 \theta \ d \theta$ and $v = \sec \theta$. Let $A = \displaystyle { \int \tan \theta \sec \theta \tan \theta \ d \theta }$. Then $$A = \displaystyle { \int \tan \theta \sec \theta \tan \theta \ d \theta } = \sec \theta \tan \theta - \displaystyle { \int \sec \theta \sec^{2} \theta \ d \theta }$$ (Recall that $\sec^2 \theta = 1 + \tan^2 \theta$.) $$= \displaystyle { \sec \theta \tan \theta - \int \sec \theta (1 + \tan^{2} \theta) \ d \theta }$$ $$= \displaystyle { \sec \theta \tan \theta - \int \sec \theta \ d \theta - \int \sec \theta \tan^{2} \theta \ d \theta }$$ (Recall that $\displaystyle { \int \sec \theta \ d \theta } = \ln\Big| \sec \theta + \tan \theta \Big| + C$.) $$= \displaystyle { \sec \theta \tan \theta - \ln\Big| \sec \theta + \tan \theta \Big| - \int \tan \theta \sec \theta \tan \theta) \ d \theta }$$ $$= \displaystyle { \sec \theta \tan \theta - \ln\Big| \sec \theta + \tan \theta \Big| - A }$$ Then $$2A = \sec \theta \tan \theta - \ln\Big| \sec \theta + \tan \theta \Big| + C$$ so that $$\displaystyle { A = \int \tan \theta \sec \theta \tan \theta \ d \theta = \frac{1}{2} \sec \theta \tan \theta - \frac{1}{2} \ln \Big|\sec \theta + \tan \theta \Big| + C }$$ and $$\displaystyle { \int { \sqrt{ {x-1} \over x } } \ dx = \displaystyle { 2 \int \tan \theta \ \sec \theta \tan \theta \ d \theta } = 2 \Big( \frac{1}{2} \sec \theta \tan \theta - \frac{1}{2} \ln \Big|\sec \theta + \tan \theta \Big| \Big) } + C$$ $$= \displaystyle { \sec \theta \tan \theta - \ln \Big|\sec \theta + \tan \theta \Big|} + C$$ $\Bigg($ We need to write our final answer in terms of $x$.

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

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