Integral Calculus Question 290
Question: $ \int_{{}}^{{}}{\frac{1}{(x^{2}-1)\sqrt{x^{2}+1}}}\ dx= $
Options:
A) $ \frac{1}{2\sqrt{2}}\log { \frac{\sqrt{1+x^{2}}+x\sqrt{2}}{\sqrt{1+x^{2}}-x\sqrt{2}} }+c $
B) $ \frac{1}{2\sqrt{2}}\log { \frac{\sqrt{1+x^{2}}-\sqrt{2}}{\sqrt{1+x^{2}}+\sqrt{2}} }+c $
C) $ \frac{1}{2\sqrt{2}}\log { \frac{\sqrt{1+x^{2}}-x\sqrt{2}}{\sqrt{1+x^{2}}+x\sqrt{2}} }+c $
D) None of these
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Answer:
Correct Answer: C
Solution:
Put $ x=\tan \theta \Rightarrow dx={{\sec }^{2}}\theta ,d\theta , $ then $ \int_{{}}^{{}}{\frac{dx}{(x^{2}-1)\sqrt{x^{2}+1}}=\int_{{}}^{{}}{\frac{{{\sec }^{2}}\theta ,d\theta }{({{\tan }^{2}}\theta -1)\sec \theta }}}=\int_{{}}^{{}}{\frac{\cos \theta ,d\theta }{(2{{\sin }^{2}}\theta -1)}} $ Again put $ t=\sin \theta \Rightarrow dt=\cos \theta ,d\theta , $ then it reduces to $ \int_{{}}^{{}}{\frac{dt}{(2t^{2}-1)}}=\frac{1}{2}\int_{{}}^{{}}{\frac{dt}{t^{2}-{{( \frac{1}{\sqrt{2}} )}^{2}}}}=\frac{1}{2\sqrt{2}}\log ( \frac{t-\frac{1}{\sqrt{2}}}{t+\frac{1}{\sqrt{2}}} )+c $ $ =\frac{1}{2\sqrt{2}}\log ( \frac{\sqrt{1+x^{2}}-x\sqrt{2}}{\sqrt{1+x^{2}}+x\sqrt{2}} )+c $ .
BETA
