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 \[\frac{8}{(x+2)({{x}^{2}}+4)}=\frac{1}{x+2}+\frac{-x+2}{{{x}^{2}}+4}\] \[\therefore \]\[\int{\frac{8}{(x+2)({{x}^{2}}+4)}dx=\int{\frac{1}{(x+2)}dx+\int{\frac{-x+2}{{{x}^{2}}+4}dx}}}\] \[=\int{\frac{1}{(x+2)}dx-\int{\frac{x}{{{x}^{2}}+4}dx+2\int{\frac{1}{{{x}^{2}}+4}dx}}}\] \[=\log |x+2|-\frac{1}{2}\int{\frac{1}{t}dt}+2.\frac{1}{2}{{\tan }^{-1}}\frac{x}{2}+C,\] (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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