NEED HELP WITH CALCULUS 1.... change in variable log rule
- Thread starter AndressaC
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You are given: \(\displaystyle \int\dfrac{\sqrt{x}}{\sqrt{x}-3}\,dx\)
and the following substitution is used:
\(\displaystyle u=\sqrt{x}-3\) and so we find:
\(\displaystyle du=\dfrac{1}{2\sqrt{x}}\,dx\,\therefore\,dx=2\sqrt{x}\,du\)
Now, if \(\displaystyle u=\sqrt{x}-3\) then \(\displaystyle \sqrt{x}=u+3\) then we have: \(\displaystyle dx=2(u+3)\,du\)
and so the integral is transformed to:
\(\displaystyle \int\dfrac{(u+3)}{u}\,2(u+3)\,du=2\int\dfrac{(u+3)^2}{u}\,du\)
Does that make more sense?
and the following substitution is used:
\(\displaystyle u=\sqrt{x}-3\) and so we find:
\(\displaystyle du=\dfrac{1}{2\sqrt{x}}\,dx\,\therefore\,dx=2\sqrt{x}\,du\)
Now, if \(\displaystyle u=\sqrt{x}-3\) then \(\displaystyle \sqrt{x}=u+3\) then we have: \(\displaystyle dx=2(u+3)\,du\)
and so the integral is transformed to:
\(\displaystyle \int\dfrac{(u+3)}{u}\,2(u+3)\,du=2\int\dfrac{(u+3)^2}{u}\,du\)
Does that make more sense?
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I am guessing that this is one of a set of problems asking for an evaluation of the integral of certain expressions using substitutions. So am I correct in guessing that the specific problem is to evaluate:
\(\displaystyle \int \dfrac{\sqrt{x}}{\sqrt{x} - 3}dx.\)
\(\displaystyle Let\ u = \sqrt{x} - 3 \implies \dfrac{du}{dx} = \dfrac{1}{2\sqrt{x}}\ and\ u + 3 = \sqrt{x} \implies \left(2\sqrt{x}\right)du = dx \implies 2(u + 3)du = dx.\) Make sense so far? Now substitute.
\(\displaystyle \int\dfrac{\sqrt{x}}{\sqrt{x} - 3}dx = \int \dfrac{u + 3}{u} * \{2(u + 3)\}du = \int \dfrac{2(u + 3)^2}{u}du = 2\int\dfrac{u^2 + 6u + 9}{u}du.\) Make sense now?
AndressaC,
do not deflect onto a problem such as this and call it "stupid" because of your
frustration with not knowing why/how a certain step was done. It is on you.
It's not a "stupid problem."
Leave out the word "stupid" in your sentence, and then it will be a sensible statement.