MHB What Is the Integral of sqrt(x-1)/x?

AI Thread Summary
The integral of sqrt(x-1)/x can be evaluated using substitution. By letting u = sqrt(x-1), the integral simplifies to 2∫(u^2/(u^2+1)) du. This leads to the result I = 2(u - arctan(u)) + C. After back-substituting for u, the final answer is I = 2sqrt(x-1) - 2arctan(sqrt(x-1)) + C. The solution effectively demonstrates the integration process through substitution.
MarkFL
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Here is the question:

What is the integral of sqrt(x-1)/x. Show Work.?


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

Use substitution.

I have posted a link there to this topic so the OP can see my work.
 
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Hello livinginmyownreality,

We are given to evaluate:

$$I=\int\frac{\sqrt{x-1}}{x}\,dx$$

Using the substitution:

$$u=\sqrt{x-1}\,\therefore\,du=\frac{1}{2\sqrt{x-1}}\,dx\,therefore\,dx=2u\,du$$

$$u^2=x-1\implies x=u^2+1$$

And so we obtain:

$$I=2\int\frac{u^2}{u^2+1}\,du=2\int\frac{u^2+1-1}{u^2+1}\,du=2\int 1-\frac{1}{u^2+1}\,du$$

From this we obtain:

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

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

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