# Integration of Rational Expression - Alternative Method

• Aug 5th 2008, 07:02 AM
scherz0
Integration of Rational Expression - Alternative Method
Hello everyone,

I integrated the following rational expression correctly, but one of my steps differ from that one of the book. Could anyone tell me if what I did is still correct?

Thank you!

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1. Integrate: $\int \frac{\sqrt{x - 4} + x}{x - 4} dx$

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$\int \frac{\sqrt{x - 4} + x}{x - 4} dx$

= $\int (x - 4)^{-1/2} dx + \int \frac{x}{x - 4}dx$

Let $u = x - 4$ and $du = dx$. Therefore, $x = u + 4$.

= $\int u^{-1/2} du + \int \frac{u + 4}{u}du$

= $2\sqrt{x - 4} + \int \frac{u}{u} du + \int \frac{4}{u} du$

= $2\sqrt{x - 4} + x + \ln(x - 4)^4 + C$

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However, my practice workbook integrated $\int \frac{u}{u} du$ as $u$, and then substituted $x - 4$ back into the equation.

So according to the workbook, its last step would be:

= $2\sqrt{x - 4} + u + \ln(x - 4)^4 + C$

= $2\sqrt{x - 4} + (x - 4) + \ln(x - 4)^4 + C$

= $2\sqrt{x - 4} + x + \ln(x - 4)^4 + C$
• Aug 5th 2008, 07:24 AM
Jameson
I think the way you did it kind of implies that $\int \frac{u}{u}du=x$ because of the way you kept your terms in order. The book's method is more correct in my opinion as it notes that the above integral doesn't equal x, but u + C, then back-substitutes x-4 for u and combines all the constants into one C.
• Aug 5th 2008, 08:48 AM
Soroban
Hello, scherz0!

What you did is correct.
. . I would not have simplifed first as you did . . .

We are given: . $\int \frac{\sqrt{x - 4} + x}{x - 4}\,dx$

Let: $\sqrt{x-4} \:=\: u \quad\Rightarrow\quad x-4\:=\:u^2 \quad\Rightarrow\quad x \:=\: u^2+4 \quad\Rightarrow\quad dx \:=\:2u\,du$

Substitute: . $\int \frac{u + (u^2+4)}{u^2}\,(2u\,du) \;\;=\;\;2\int\frac{u^2+u-4}{u}\,du \;\;=\;\; 2\int\left(u + 1 - \frac{4}{u}\right)\,du$

. . . . . . $= \;\;2\left(\frac{u^2}{2} + u - 4\ln|u|\right) + C \;\;=\;\;u^2 + 2u - 8 \ln|u| + C$

Back-substitute: . $(x - 4) + 2\sqrt{x-4} - 8\ln\left(\sqrt{x-4}\right) + C$

. . . . . . . . . . . $=\;\;x - 4 + 2\sqrt{x-4} - 8\ln(x-4)^{\frac{1}{2}} + C$

. . . . . . . . . . . $= \;\;x + 2\sqrt{x-4} - 4\ln(x-4) + \underbrace{C - 4}_{\text{a constant}}$

. . . . . . . . . . . $= \;\;x + 2\sqrt{x-4} - 4\ln(x-4) + C$