A) \[\frac{x}{1+x}+\log (1+x)\]
B) \[\frac{x}{1-x}+\log (1+x)\]
C) \[-\frac{x}{1-x}+\log (1+x)\]
D) \[\frac{x}{1-x}+\log (1-x)\]
Correct Answer: D
Solution :
| [d] \[\frac{1}{2}{{x}^{2}}+\frac{2}{3}{{x}^{3}}+\frac{3}{4}{{x}^{4}}+\frac{4}{5}{{x}^{5}}+.....\] |
| \[=\left( 1-\frac{1}{2} \right){{x}^{2}}+\left( 1-\frac{1}{3} \right){{x}^{3}}+\left( 1-\frac{1}{4} \right){{x}^{4}}+\] |
| \[\left( 1-\frac{1}{5} \right){{x}^{5}}+...\] |
| \[=({{x}^{2}}+{{x}^{3}}+{{x}^{4}}+{{x}^{5}}+......)+\] |
| \[\left( -\frac{{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4}-\frac{{{x}^{5}}}{5}...... \right)\] |
| \[=(x+{{x}^{2}}+{{x}^{3}}+...)+\left( -\frac{{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4}-\frac{{{x}^{5}}}{5}...... \right)\] |
| \[=\frac{x}{1-x}+{{\log }_{e}}(1-x)\] |
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