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If \(f(x)=\int \frac{5 x^{8}+7 x^{6}}{\left(x^{2}+1+2 x^{7}\right)^{2}} d x,(x \geq 0),\) and \(f(0)=0,\) then the value of \(f(1)\) is( )
A. \(-\frac{1}{2}\)
B. \(-\frac{1}{2}\)
C. \(\frac{1}{4}\)
D. \(\frac{1}{2}\)
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If \(f(x)=\int \frac{5 x^{8}+7 x^{6}}{\left(x^{2}+1+2 x^{7}\right)^{2}} d x,(x \geq 0),\) and \(f(0)=0,\) then the value of \(f(1)\) is( )
A. \(-\frac{1}{2}\)
B. \(-\frac{1}{2}\)
C. \(\frac{1}{4}\)
D. \(\frac{1}{2}\)

Answer

C
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Solution

We have, \(f(x)=\int \frac{5 x^{8}+7 x^{6}}{\left(x^{2}+1+2 x^{7}\right)^{2}} d x\)
\(=\int \frac{5\left(\frac{x^{8}}{x^{14}}\right)+7\left(\frac{x^{6}}{x^{14}}\right)}{\left(\frac{x^{2}}{x^{7}}+\frac{1}{x^{7}}+\frac{2 x^{7}}{x^{7}}\right)^{2}} d x\)
(dividing both numerator and denominator by \(x^{14}\) )
\(=\int \frac{5 x^{-6}+7 x^{-8}}{\left(x^{-5}+x^{-7}+2\right)^{2}} d x\)
Let \(x^{-5}+x^{-7}+2=t\)
\(\Rightarrow\left(-5 x^{-6}-7 x^{-8}\right) d x=d t\)
\(\Rightarrow \quad\left(5 x^{-6}+7 x^{-8}\right) d x=-d t\)
\(\therefore f(x)=\int-\frac{d t}{t^{2}}=-\int t^{-2} d t\)
\(=-\frac{t^{-2+1}}{-2+1}+C=-\frac{t^{-1}}{-1}+C=\frac{1}{t}+C\)
\(=\frac{1}{x^{-5}+x^{-7}+2}+C=\frac{x^{7}}{2 x^{7}+x^{2}+1}+C\)
\(\because f(0)=0\)
\(\therefore 0=\frac{0}{0+0+1}+C \Rightarrow C=0\)
\(\therefore f(x)=\frac{x^{7}}{2 x^{7}+x^{2}+1}\)
\(\Rightarrow f(1)=\frac{1}{2(1)^{7}+1^{2}+1}=\frac{1}{4}\)
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