- Thread starter
- #1
Albert
Well-known member
- Jan 25, 2013
- 1,225
$ \begin{align*} \int_{1}^{2}\dfrac {dx}{(x^2-2x+4)^{3/2}}\end{align*}$
Last edited:
When problems are posted here in our Challenge Questions and Puzzles sub-forum, we are to assume the OP is not looking for hints and suggestions, but rather has already solved the problem, and found the problem interesting enough to post here as a challenge to our members.As a first step I would complete the square.![]()
\(\displaystyle \displaystyle \begin{align*} \int_1^2{\frac{dx}{\left( x^2 - 2x + 4 \right) ^{\frac{3}{2}} }} &= \int_1^2{\frac{dx}{\left[ \left( x - 1 \right) ^2 + 3 \right] ^{\frac{3}{2}} }} \end{align*}\)$\int_{1}^{2}\dfrac {dx}{(x^2-2x+4)^{3/2}}$
perfect ! you got it,your solution is really\(\displaystyle \displaystyle \begin{align*} \int_1^2{\frac{dx}{\left( x^2 - 2x + 4 \right) ^{\frac{3}{2}} }} &= \int_1^2{\frac{dx}{\left[ \left( x - 1 \right) ^2 + 3 \right] ^{\frac{3}{2}} }} \end{align*}\)
Now make the substitution \(\displaystyle \displaystyle \begin{align*} x - 1 = \sqrt{3}\,\tan{(\theta)} \implies dx = \sqrt{3}\,\sec^2{(\theta)}\,d\theta \end{align*}\) and noting that when \(\displaystyle \displaystyle \begin{align*} x = 1, \, \theta = 0 \end{align*}\) and when \(\displaystyle \displaystyle \begin{align*} x = 2, \, \theta = \frac{\pi}{6} \end{align*}\)
--------