A) \[\frac{1}{3}\]
B) \[\frac{1}{2}\]
C) \[\frac{1}{4}\]
D) \[2\]
E) \[1\]
Correct Answer: C
Solution :
Let \[x+2=A\frac{d}{dx}(2{{x}^{2}}+6x+5)+B\] \[\Rightarrow \] \[x+2=A(4x+6)+B\] \[\Rightarrow \] \[x+2=4Ax+6A+B\] \[\Rightarrow \] \[4A=1\] \[\Rightarrow \] \[A=\frac{1}{4}\] \[6A+B=2\] \[\Rightarrow \] \[B=\frac{1}{2}\] \[\therefore \]\[\int{\frac{x+2}{2{{x}^{2}}+6x+5}}dx\] \[\int{\frac{\left( \frac{1}{4}(4x+6)+\frac{1}{2} \right)}{2{{x}^{2}}+6x+5}}dx\] \[=\frac{1}{4}\int{\frac{4x+6}{2{{x}^{2}}+6x+5}}dx+\frac{1}{2}\int{\frac{2}{2{{x}^{2}}+6x+5}}dx\] Comparing it with\[\int{\frac{x+2}{2{{x}^{2}}+6x+5}}dx\] \[=p\int{\frac{4x+6}{2{{x}^{2}}+6x+5}}dx+\frac{1}{2}\int{\frac{dx}{2{{x}^{2}}+6x+15}}\] \[\Rightarrow \] \[p=\frac{1}{4}\] Alternate \[p(4x+6)+\frac{1}{2}=x+2\] \[\Rightarrow \] \[p(4x+6)=x+\frac{3}{2}\] \[\Rightarrow \] \[p=\frac{x+\frac{3}{2}}{4x+6}=\frac{1}{4}\]You need to login to perform this action.
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