A) \[A=1,\text{ }B=1\]
B) \[A=1,\text{ }B=2\]
C) \[A=2,B=2\]
D) \[A=2,B=-2\]
E) \[A=-2,B=-2\]
Correct Answer: D
Solution :
Let \[I=\int{\frac{\sqrt{x}}{x+1}dx}=\int{\frac{\sqrt{x}dx}{{{(\sqrt{x})}^{2}}+1}}\] Let \[\sqrt{x}=t\] \[\Rightarrow \] \[\frac{1}{2\sqrt{x}}dx=dt\] \[\Rightarrow \] \[dx=2\sqrt{x}dt\] \[\therefore \] \[I=\int{\frac{2{{t}^{2}}dt}{{{t}^{2}}+1}}\] \[=2\left[ \int{\frac{{{t}^{2}}+1}{{{t}^{2}}+1}dt-\int{\frac{1}{{{t}^{2}}+1}dt}} \right]\] \[=2t-2{{\tan }^{-1}}({{t}^{2}}+1)+c\] \[=2\sqrt{x}-2{{\tan }^{-1}}(x+1)+c\] But \[I=A\sqrt{x}+B{{\tan }^{-1}}(x+1)+c\] \[\Rightarrow \] \[A=2,B=-2\]
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