(A) : Given, \(\int \frac{7 x^{8}+8 x^{7}}{\left(1+x+x^{8}\right)^{2}} d x=f(x)+C\) So, \(I=\int \frac{7 x^{8}+8 x^{7}}{\left(x^{8}+x+1\right)^{2}} d x\) \(=\int \frac{7 x^{8}+8 x^{7}}{\left(\frac{x^{8}+x+1}{x^{8}}\right)^{2} \cdot x^{16}}=\int \frac{\frac{7}{x^{8}}+\frac{8}{x^{9}}}{\left(1+\frac{1}{x^{7}}+\frac{1}{x^{8}}\right)^{2}} d x\) Put \(1+\frac{1}{\mathrm{x}^{7}}+\frac{1}{\mathrm{x}^{8}}=\mathrm{t}\) \(\Rightarrow-\frac{7}{x^{8}}-\frac{8}{x^{9}} d x=d t\) \(\Rightarrow\left(\frac{7}{x^{8}}+\frac{8}{x^{9}}\right) d x=-d t\) \(I=-\int \frac{d t}{t^{2}}=\frac{1}{t}+C\) \(=\frac{1}{1+\frac{1}{x^{7}}+\frac{1}{x_{8}^{8}}}+C=\frac{x^{8}}{x^{8}+x+1}+C\) So, \(\quad f(x)=\frac{x^{8}}{x^{8}+x+1}\)
AP EAMCET-2010
Integral Calculus
86139
If \(f_{n}(x)=\log \log \log \ldots . \log x(\log\) is repeated \(n-\) times), then \(\int\left(\mathrm{xf}_{1}(\mathrm{x}) \mathrm{f}_{2}(\mathrm{x}) \ldots \ldots \ldots \mathrm{f}_{\mathrm{n}}(\mathrm{x})\right)^{-1} \mathrm{dx}\) is equal to
(A) : Given, \(\int \frac{7 x^{8}+8 x^{7}}{\left(1+x+x^{8}\right)^{2}} d x=f(x)+C\) So, \(I=\int \frac{7 x^{8}+8 x^{7}}{\left(x^{8}+x+1\right)^{2}} d x\) \(=\int \frac{7 x^{8}+8 x^{7}}{\left(\frac{x^{8}+x+1}{x^{8}}\right)^{2} \cdot x^{16}}=\int \frac{\frac{7}{x^{8}}+\frac{8}{x^{9}}}{\left(1+\frac{1}{x^{7}}+\frac{1}{x^{8}}\right)^{2}} d x\) Put \(1+\frac{1}{\mathrm{x}^{7}}+\frac{1}{\mathrm{x}^{8}}=\mathrm{t}\) \(\Rightarrow-\frac{7}{x^{8}}-\frac{8}{x^{9}} d x=d t\) \(\Rightarrow\left(\frac{7}{x^{8}}+\frac{8}{x^{9}}\right) d x=-d t\) \(I=-\int \frac{d t}{t^{2}}=\frac{1}{t}+C\) \(=\frac{1}{1+\frac{1}{x^{7}}+\frac{1}{x_{8}^{8}}}+C=\frac{x^{8}}{x^{8}+x+1}+C\) So, \(\quad f(x)=\frac{x^{8}}{x^{8}+x+1}\)
AP EAMCET-2010
Integral Calculus
86139
If \(f_{n}(x)=\log \log \log \ldots . \log x(\log\) is repeated \(n-\) times), then \(\int\left(\mathrm{xf}_{1}(\mathrm{x}) \mathrm{f}_{2}(\mathrm{x}) \ldots \ldots \ldots \mathrm{f}_{\mathrm{n}}(\mathrm{x})\right)^{-1} \mathrm{dx}\) is equal to
(A) : Given, \(\int \frac{7 x^{8}+8 x^{7}}{\left(1+x+x^{8}\right)^{2}} d x=f(x)+C\) So, \(I=\int \frac{7 x^{8}+8 x^{7}}{\left(x^{8}+x+1\right)^{2}} d x\) \(=\int \frac{7 x^{8}+8 x^{7}}{\left(\frac{x^{8}+x+1}{x^{8}}\right)^{2} \cdot x^{16}}=\int \frac{\frac{7}{x^{8}}+\frac{8}{x^{9}}}{\left(1+\frac{1}{x^{7}}+\frac{1}{x^{8}}\right)^{2}} d x\) Put \(1+\frac{1}{\mathrm{x}^{7}}+\frac{1}{\mathrm{x}^{8}}=\mathrm{t}\) \(\Rightarrow-\frac{7}{x^{8}}-\frac{8}{x^{9}} d x=d t\) \(\Rightarrow\left(\frac{7}{x^{8}}+\frac{8}{x^{9}}\right) d x=-d t\) \(I=-\int \frac{d t}{t^{2}}=\frac{1}{t}+C\) \(=\frac{1}{1+\frac{1}{x^{7}}+\frac{1}{x_{8}^{8}}}+C=\frac{x^{8}}{x^{8}+x+1}+C\) So, \(\quad f(x)=\frac{x^{8}}{x^{8}+x+1}\)
AP EAMCET-2010
Integral Calculus
86139
If \(f_{n}(x)=\log \log \log \ldots . \log x(\log\) is repeated \(n-\) times), then \(\int\left(\mathrm{xf}_{1}(\mathrm{x}) \mathrm{f}_{2}(\mathrm{x}) \ldots \ldots \ldots \mathrm{f}_{\mathrm{n}}(\mathrm{x})\right)^{-1} \mathrm{dx}\) is equal to
(A) : Given, \(\int \frac{7 x^{8}+8 x^{7}}{\left(1+x+x^{8}\right)^{2}} d x=f(x)+C\) So, \(I=\int \frac{7 x^{8}+8 x^{7}}{\left(x^{8}+x+1\right)^{2}} d x\) \(=\int \frac{7 x^{8}+8 x^{7}}{\left(\frac{x^{8}+x+1}{x^{8}}\right)^{2} \cdot x^{16}}=\int \frac{\frac{7}{x^{8}}+\frac{8}{x^{9}}}{\left(1+\frac{1}{x^{7}}+\frac{1}{x^{8}}\right)^{2}} d x\) Put \(1+\frac{1}{\mathrm{x}^{7}}+\frac{1}{\mathrm{x}^{8}}=\mathrm{t}\) \(\Rightarrow-\frac{7}{x^{8}}-\frac{8}{x^{9}} d x=d t\) \(\Rightarrow\left(\frac{7}{x^{8}}+\frac{8}{x^{9}}\right) d x=-d t\) \(I=-\int \frac{d t}{t^{2}}=\frac{1}{t}+C\) \(=\frac{1}{1+\frac{1}{x^{7}}+\frac{1}{x_{8}^{8}}}+C=\frac{x^{8}}{x^{8}+x+1}+C\) So, \(\quad f(x)=\frac{x^{8}}{x^{8}+x+1}\)
AP EAMCET-2010
Integral Calculus
86139
If \(f_{n}(x)=\log \log \log \ldots . \log x(\log\) is repeated \(n-\) times), then \(\int\left(\mathrm{xf}_{1}(\mathrm{x}) \mathrm{f}_{2}(\mathrm{x}) \ldots \ldots \ldots \mathrm{f}_{\mathrm{n}}(\mathrm{x})\right)^{-1} \mathrm{dx}\) is equal to
(A) : Given, \(\int \frac{7 x^{8}+8 x^{7}}{\left(1+x+x^{8}\right)^{2}} d x=f(x)+C\) So, \(I=\int \frac{7 x^{8}+8 x^{7}}{\left(x^{8}+x+1\right)^{2}} d x\) \(=\int \frac{7 x^{8}+8 x^{7}}{\left(\frac{x^{8}+x+1}{x^{8}}\right)^{2} \cdot x^{16}}=\int \frac{\frac{7}{x^{8}}+\frac{8}{x^{9}}}{\left(1+\frac{1}{x^{7}}+\frac{1}{x^{8}}\right)^{2}} d x\) Put \(1+\frac{1}{\mathrm{x}^{7}}+\frac{1}{\mathrm{x}^{8}}=\mathrm{t}\) \(\Rightarrow-\frac{7}{x^{8}}-\frac{8}{x^{9}} d x=d t\) \(\Rightarrow\left(\frac{7}{x^{8}}+\frac{8}{x^{9}}\right) d x=-d t\) \(I=-\int \frac{d t}{t^{2}}=\frac{1}{t}+C\) \(=\frac{1}{1+\frac{1}{x^{7}}+\frac{1}{x_{8}^{8}}}+C=\frac{x^{8}}{x^{8}+x+1}+C\) So, \(\quad f(x)=\frac{x^{8}}{x^{8}+x+1}\)
AP EAMCET-2010
Integral Calculus
86139
If \(f_{n}(x)=\log \log \log \ldots . \log x(\log\) is repeated \(n-\) times), then \(\int\left(\mathrm{xf}_{1}(\mathrm{x}) \mathrm{f}_{2}(\mathrm{x}) \ldots \ldots \ldots \mathrm{f}_{\mathrm{n}}(\mathrm{x})\right)^{-1} \mathrm{dx}\) is equal to