Hi guys the problem + attempt to solve it is attached to the image,
how can i finish it?
http://img141.imageshack.us/i/85802191.jpg/
how can i finish it?
http://img141.imageshack.us/i/85802191.jpg/
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Determine the length of the curve : y = ln I x^2 -1 I
- Thread starter Riazy
- Start date
You got to \(\displaystyle \sqrt{\frac{4x^{2}}{(x^{2}-1)^{2}}+1}=\frac{x^{2}+1}{x^{2}-1}\)
\(\displaystyle \int_{0}^{\frac{1}{2}}\frac{x^{2}+1}{x^{2}-1}dx=\int_{0}^{\frac{1}{2}}\frac{-1}{x+1}dx+\int_{0}^{\frac{1}{2}}\frac{1}{x-1}dx+\int_{0}^{\frac{1}{2}}dx\)
Now, it is easy to finish?.
\(\displaystyle \int_{0}^{\frac{1}{2}}\frac{x^{2}+1}{x^{2}-1}dx=\int_{0}^{\frac{1}{2}}\frac{-1}{x+1}dx+\int_{0}^{\frac{1}{2}}\frac{1}{x-1}dx+\int_{0}^{\frac{1}{2}}dx\)
Now, it is easy to finish?.
It is just like when you added fractions back in algebra.
\(\displaystyle \frac{4x^{2}}{(x^{2}-1)^{2}}+\underbrace{\frac{(x^{2}-1)^{2}}{(x^{2}-1)^{2}}}_{\text{1}}\)
\(\displaystyle \frac{4x^{2}+(x^{2}-1)^{2}}{(x^{2}-1)^{2}}\)
The top reduces to \(\displaystyle (x^{2}+1)^{2}\),
Then, take the sqaure root and you have the result.
\(\displaystyle \frac{4x^{2}}{(x^{2}-1)^{2}}+\underbrace{\frac{(x^{2}-1)^{2}}{(x^{2}-1)^{2}}}_{\text{1}}\)
\(\displaystyle \frac{4x^{2}+(x^{2}-1)^{2}}{(x^{2}-1)^{2}}\)
The top reduces to \(\displaystyle (x^{2}+1)^{2}\),
Then, take the sqaure root and you have the result.