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I have posted a link there to this topic so the OP can see my work.What is the integral of sqrt(x-1)/x. Show Work.?

Answer should be 2sqrt(x-1) - 2arctan*sqrt(x-1) + C

Use substitution.

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- Thread starter
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- #1

I have posted a link there to this topic so the OP can see my work.What is the integral of sqrt(x-1)/x. Show Work.?

Answer should be 2sqrt(x-1) - 2arctan*sqrt(x-1) + C

Use substitution.

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We are given to evaluate:

\(\displaystyle I=\int\frac{\sqrt{x-1}}{x}\,dx\)

Using the substitution:

\(\displaystyle u=\sqrt{x-1}\,\therefore\,du=\frac{1}{2\sqrt{x-1}}\,dx\,therefore\,dx=2u\,du\)

\(\displaystyle u^2=x-1\implies x=u^2+1\)

And so we obtain:

\(\displaystyle I=2\int\frac{u^2}{u^2+1}\,du=2\int\frac{u^2+1-1}{u^2+1}\,du=2\int 1-\frac{1}{u^2+1}\,du\)

From this we obtain:

\(\displaystyle I=2\left(u-\tan^{-1}(u) \right)+C\)

Back-substituting for $u$, and distributing the $2$, we get:

\(\displaystyle I=2\sqrt{x-1}-2\tan^{-1}\left(\sqrt{x-1} \right)+C\)