Integral Calculus Question 26
Question: $ \int_{{}}^{{}}{\frac{1}{(x-1)(x^{2}+1)}dx}= $
[Roorkee 1984]
Options:
A) $\frac{1}{2} \log \left|x-1 \right|-\frac{1}{4} \log \left| x^2+1 \right| -\frac{1}{2} \tan ^{-1} x+C$
B) $\frac{1}{2} \log \left|x-1\right| +\frac{1}{4} \log \left|x^2+1\right|-\frac{1}{2} \tan ^{-1} x+C$
C) $\frac{1}{2} \log \left|x-1\right| -\frac{1}{2} \log \left|x^2+1\right|-\frac{1}{2} \tan ^{-1} x+C$
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
We have $ \frac{1}{(x-1)(x^{2}+1)}=\frac{A}{(x-1)}+\frac{Bx+C}{(x^{2}+1)} $
$ \Rightarrow 1=A(x^{2}+1)+(Bx+C)(x-1) $
If $ x=1, $ then $ A=\frac{1}{2} $ …..(i)
$ A-C=1\Rightarrow C=-\frac{1}{2} $ …..(ii) $ A+B=0\Rightarrow B=-\frac{1}{2} $ …..(iii)
Putting these values, we get $ \frac{1}{(x-1)(x^{2}+1)}=\frac{1}{2}.\frac{1}{(x-1)}-\frac{x+1}{2(x^{2}+1)} $
Hence $ \int_{{}}^{{}}{\frac{1}{(x-1)(x^{2}+1)}}dx=\frac{1}{2}\int_{{}}^{{}}{\frac{dx}{(x-1)}-\frac{1}{2}},\int_{{}}^{{}}{\frac{x+1}{x^{2}+1}}dx $ $ =\frac{1}{2}\log (x-1)-\frac{1}{4}\log (x^{2}+1)-\frac{1}{2}{{\tan }^{-1}}x+c. $
BETA
