Question

calc 2



\[\int{\frac{1}{\sqrt{{{x}^{2}}-2x+10}}}\]

Answer 1

\[\int\limits \frac{1}{\sqrt{x^2-2x+10}} dx=\int\limits \frac{dx}{\sqrt{x^2-2x+1+9}}\]
\[=\int\limits \frac{dx}{\sqrt{(x-1)^2+3^2}}\]
Let (x-1) = 3 tan θ
therefore\[ dx = 3\sec^2 \theta d \theta \]
Hence \[\int\limits\limits \frac{dx}{\sqrt{(x-1)^2+3^2}} =\int\limits\limits \frac{1}{\sqrt{(3 \tan \theta)^2+3^2}} \times 3 \sec^2 \theta d \theta\]
\[ =\int\limits\limits \frac{3 \sec^2 \theta d \theta}{\sqrt{3^2 \tan^2 \theta+3^2}} \]
\[=\int\limits\limits \frac{3 \sec^2 \theta d \theta}{3\sqrt{ \tan^2 \theta+1}} \]
\[=\int\limits\limits \frac{\sec^2 \theta d \theta}{\sqrt{ \sec^2 \theta}} \]
\[=\int\limits\limits \frac{\sec^2 \theta d \theta}{ \sec \theta} \]
\[=\int\limits \sec \theta d \theta\]
=log |sec θ + tan θ| +c
since 3 tan θ =(x-1) [As we have assumed already}
Therefore \[\tan θ = \frac{(x-1)}{3}\]
But \[\sec^2 \theta = \tan^2 \theta +1= [\frac{(x-1)}{3}]^2 +1= \frac{(x-1)^2}{3^2} +1\]
\[= \frac{(x-1)^2}{9} +1= \frac{(x-1)^2+9}{9}\]
\[\rightarrow \sec \theta = \sqrt \frac{(x-1)^2+9}{9}= \frac{\sqrt{x^2-2x+1+9}}{3}\]
\[\rightarrow \sec \theta = \frac{\sqrt{x^2-2x+10}}{3}\]
\[\int\limits\limits \frac{1}{\sqrt{x^2-2x+10}} dx= \log| \sec \theta + \tan \theta | +c \]
\[= \log |\frac{\sqrt{x^2-2x+10}}{3}+\frac{x-1}{3}= \log |\frac{(x-1)+\sqrt{x^2-2x+10}}{3}|+C\] is the required answer of the given integration.

@Doll12

Answer 2

@Doll12 Your integration problem has been integrated.. Take a look..

Answer 3

How did you know how to do this https://www.dropbox.com/s/28pkc1k1juz2xgy/Screenshot%202013-10-26%2021.23.02.jpg