A) \[2\sqrt{1+\sqrt{x}}-2\log \left( \frac{\sqrt{1+\sqrt{x}}-1}{\sqrt{1+\sqrt{x}}+1} \right)+c\]
B) \[4\sqrt{1+\sqrt{x}}-2\log \left( \frac{\sqrt{1+\sqrt{x}}-1}{\sqrt{1+\sqrt{x}}+1} \right)+c\]
C) \[4\sqrt{1+\sqrt{x}}-2\log \left( \frac{\sqrt{1+\sqrt{x}}-1}{\sqrt{1+\sqrt{x}}+1} \right)+c\]
D) \[2\sqrt{1+\sqrt{x}}+2\log \left( \frac{\sqrt{1+\sqrt{x}}-1}{\sqrt{1+\sqrt{x}}+1} \right)+c\]
Correct Answer: B
Solution :
[b] \[\because I=\int{\sqrt{\frac{1+\sqrt{x}}{x}}}.dx\] Let \[1+\sqrt{x}={{z}^{2}}\] \[\Rightarrow \frac{1}{2x}.dx=2z.dz\] Now, \[I=\int{\frac{x.4z\sqrt{x}.dz}{x}}=4\int{\frac{{{z}^{2}}}{\sqrt{x}}.dz}=4\int{\frac{{{z}^{2}}}{{{z}^{2}}-1}.dz}\] \[=4\int{\frac{\left( {{z}^{2}}-1+1 \right).dz}{{{z}^{2}}-1}}=4\int{\left( 1+\frac{1}{{{z}^{2}}-1} \right)}.dz\] \[=4.\left[ z+\frac{1}{2}.\log \frac{z-1}{z+1} \right]+c=4.\left[ \sqrt{1+\sqrt{x}}+\frac{1}{2}.\log \left( \frac{\sqrt{1+\sqrt{x}}-1}{\sqrt{1+\sqrt{x}}+1} \right) \right]+c\] \[=4\sqrt{1+\sqrt{x}}+2.\log \left( \frac{\sqrt{1+\sqrt{x}}-1}{\sqrt{1+\sqrt{x}}+1} \right)+c\] Hence, option [b] is correct.
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