A) \[\frac{1}{3}(x+1)\]
B) \[\frac{1}{3}(x+4)\]
C) \[\frac{2}{3}(x+2)\]
D) \[\frac{2}{3}(x-4)\]
Correct Answer: B
Solution :
Let\[I=\int_{{}}^{{}}{\frac{x+1}{\sqrt{2x-1}}}dx\] Put\[2x-1={{t}^{2}}\Rightarrow 2dx=2t\,dt\] \[\therefore \]\[I=\int_{{}}^{{}}{\frac{\frac{{{t}^{2}}+1}{2}+1}{t}t\,dt}=\frac{1}{2}\int_{{}}^{{}}{({{t}^{2}}+3)}dt\] \[=\frac{1}{2}\left( \frac{{{t}^{3}}}{3}+3t \right)+C=\frac{t}{6}({{t}^{2}}+9)+C\] \[=\frac{\sqrt{2x-1}}{6}(2x-1+9)+C=\sqrt{2x-1}\left( \frac{x+4}{3} \right)+C\] \[\therefore \]\[f(x)=\frac{x+4}{3}\]
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