A) \[xlog(x+\sqrt{{{x}^{2}}+{{a}^{2}}})+\sqrt{{{x}^{2}}+{{a}^{2}}}+c\]
B) \[xlog(x+\sqrt{{{x}^{2}}+{{a}^{2}}})-\sqrt{{{x}^{2}}+{{a}^{2}}}+c\]
C) \[x.log(x+\sqrt{{{x}^{2}}+{{a}^{2}}})-\sqrt{{{x}^{2}}+{{a}^{2}}}+c\]
D) None of these
Correct Answer: C
Solution :
[c] \[\because I=\int{\log \left( x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right)}=\int{1\times \log \left( x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right)}\] \[=\log \left( x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right).x-\int{\frac{1}{x+\sqrt{{{x}^{2}}+{{a}^{2}}}}\times \left( 1+\frac{1\times 2x}{2\sqrt{{{x}^{2}}+{{a}^{2}}}} \right)}.x\times dx\] \[=x.\log (x+\sqrt{{{x}^{2}}+{{a}^{2}}})-\int{\frac{x}{x+\sqrt{{{x}^{2}}+{{a}^{2}}}}.\left( 1+\frac{x}{\sqrt{{{x}^{2}}+{{a}^{2}}}} \right)}.dx\]\[=x.\log (x+\sqrt{{{x}^{2}}+{{a}^{2}}})-\int{\frac{x}{\sqrt{{{x}^{2}}+{{a}^{2}}}}.dx}\] \[=x.\log (x+\sqrt{{{x}^{2}}+{{a}^{2}}})-\sqrt{{{x}^{2}}+{{a}^{2}}}+c\] \[\left[ \because \int{\frac{1}{\sqrt{2}}.dz=\frac{1}{2}.\sqrt{z}+c.} \right]\] Hence, option [c] is correct.
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