A) \[\log |x+2|-\frac{1}{2}\log ({{x}^{2}}+4)+{{\tan }^{-1}}\frac{x}{2}+C\]
B) \[\log |x+2|-\frac{1}{2}\log ({{x}^{2}}+4)+{{\sin }^{-1}}\frac{x}{2}+C\]
C) \[\log |x+2|-\frac{1}{2}\log ({{x}^{2}}+4)+{{\cos }^{-1}}\frac{x}{2}+C\]
D) \[\log |x+2|-\frac{1}{2}\log ({{x}^{3}}+4)+{{\tan }^{-1}}\frac{x}{2}+C\]
Correct Answer: A
Solution :
Let\[\frac{8}{(x+2)({{x}^{2}}+4)}=\frac{A}{x+2}+\frac{Bx+C}{{{x}^{2}}+4}\] ?(i) Then\[8=A({{x}^{2}}+4)+(Bx+C)(x+2)\] ?(ii) Putting \[x+2=0\]i.e. \[x=-2\]in (ii), we get\[8=8A\] \[\Rightarrow A=1\] Putting x = 0 and 1 respectively in (ii), we get \[8=4A+2C\]and\[8=5A+3B+3C\] Solving these equation, we obtain \[A=1,C=2,B=-1\] Substituting the values of A, B and C in (i), we obtain \[\frac{8}{(x+2)({{x}^{2}}+4)}=\frac{1}{x+2}+\frac{-x+2}{{{x}^{2}}+4}\] \[\therefore \]\[\int_{{}}^{{}}{\frac{8}{(x+2)({{x}^{2}}+4)}}dx=\int_{{}}^{{}}{\frac{1}{(x+2)}}dx+\int\limits_{{}}^{{}}{\frac{-x+2}{{{x}^{2}}+4}}dx\] \[=\int_{{}}^{{}}{\frac{1}{(x+2)}}dx-\int\limits_{{}}^{{}}{\frac{x}{{{x}^{2}}+4}}dx+2\int_{{}}^{{}}{\frac{1}{{{x}^{2}}+4}}dx\] \[=\log |x+2|-\frac{1}{2}\int_{{}}^{{}}{\frac{1}{t}dt+2.\frac{1}{2}{{\tan }^{-1}}\frac{x}{2}}+C,\] (where\[t={{x}^{2}}+4\]) \[=\log |x+2|-\frac{1}{2}\log t+{{\tan }^{-1}}\frac{x}{2}+C\] \[=\log |x+2|-\frac{1}{2}\log ({{x}^{2}}+4)+{{\tan }^{-1}}\frac{x}{2}+C\]
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