A) \[-\frac{1}{2}\]
B) \[-\frac{1}{4}\]
C) \[\frac{1}{4}\]
D) \[\frac{1}{2}\]
Correct Answer: C
Solution :
\[f(x)\,=\,\int{\frac{5{{x}^{8}}+7{{x}^{6}}}{{{({{x}^{2}}+1+2{{x}^{7}})}^{2}}}\,dx}\] \[f(x)=\int{\frac{5{{x}^{8}}+7{{x}^{6}}}{{{x}^{14}}{{({{x}^{-5}}+{{x}^{-7}}+2)}^{2}}}\,dx}\] \[f(x)\,=\,\int{\frac{5{{x}^{-6}}+7{{x}^{-8}}}{{{({{x}^{-5}}\,+{{x}^{-7}}+2)}^{2}}}\,dx}\] \[{{x}^{-}}^{5}+{{x}^{-}}^{7}+2=t\] \[\left( -5{{x}^{-}}^{6}-7{{x}^{-8}} \right)dx=dt\] \[\left( 5{{x}^{-}}^{6}+\text{ }7{{x}^{-8}} \right)dx=-dt\] \[f(x)\,\,=\,\int{\frac{-dt}{{{t}^{2}}}\,\,=\,\,\frac{1}{t}}+c\] \[f(x)\,\,=\,\frac{1}{{{x}^{-5}}+{{x}^{-7}}+2}+c\] \[f(x)\,\,=\,\frac{{{x}^{7}}}{{{x}^{2}}+1+2{{x}^{7}}}+c\] \[f\left( 0 \right)=0~~~~~\therefore \,\,\text{ }c=0\] \[f(x)\,\,=\,\,\frac{{{x}^{7}}}{({{x}^{2}}+1+2{{x}^{7}})}\] \[f(1)\,\,=\,\,\frac{1}{1+1+2}\,\,=\,\,\frac{1}{4}\]
You need to login to perform this action.
You will be redirected in
3 sec
