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 $x$.

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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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