(B) : Given, \(I=\int \frac{x^{5} d x}{\left(x^{2}+x+1\right)\left(x^{6}+1\right)\left(x^{4}-x^{3}+x-1\right)}\) \(=\int \frac{x^{5} d x}{\left(x^{6}+1\right)\left(x^{3}+1\right)(x-1)\left(x^{2}+x+1\right)}=\int \frac{x^{5} d x}{\left(x^{6}+1\right)\left(x^{6}-1\right)}\) Let, \(x^{6}=t\) \(6 x^{5} d x=d t\) So, \(=\frac{1}{6} \int \frac{\mathrm{dt}}{(\mathrm{t}+1)(\mathrm{t}-1)}\) \(=\frac{1}{12} \log _{\mathrm{e}}\left|\frac{\mathrm{t}-1}{\mathrm{t}+1}\right|+\mathrm{C}=\frac{1}{12} \log _{\mathrm{e}}\left|\frac{\mathrm{x}^{6}-1}{\mathrm{x}^{6}+1}\right|+\mathrm{C}\)
(B) : Given, \(I=\int \frac{x^{5} d x}{\left(x^{2}+x+1\right)\left(x^{6}+1\right)\left(x^{4}-x^{3}+x-1\right)}\) \(=\int \frac{x^{5} d x}{\left(x^{6}+1\right)\left(x^{3}+1\right)(x-1)\left(x^{2}+x+1\right)}=\int \frac{x^{5} d x}{\left(x^{6}+1\right)\left(x^{6}-1\right)}\) Let, \(x^{6}=t\) \(6 x^{5} d x=d t\) So, \(=\frac{1}{6} \int \frac{\mathrm{dt}}{(\mathrm{t}+1)(\mathrm{t}-1)}\) \(=\frac{1}{12} \log _{\mathrm{e}}\left|\frac{\mathrm{t}-1}{\mathrm{t}+1}\right|+\mathrm{C}=\frac{1}{12} \log _{\mathrm{e}}\left|\frac{\mathrm{x}^{6}-1}{\mathrm{x}^{6}+1}\right|+\mathrm{C}\)
(B) : Given, \(I=\int \frac{x^{5} d x}{\left(x^{2}+x+1\right)\left(x^{6}+1\right)\left(x^{4}-x^{3}+x-1\right)}\) \(=\int \frac{x^{5} d x}{\left(x^{6}+1\right)\left(x^{3}+1\right)(x-1)\left(x^{2}+x+1\right)}=\int \frac{x^{5} d x}{\left(x^{6}+1\right)\left(x^{6}-1\right)}\) Let, \(x^{6}=t\) \(6 x^{5} d x=d t\) So, \(=\frac{1}{6} \int \frac{\mathrm{dt}}{(\mathrm{t}+1)(\mathrm{t}-1)}\) \(=\frac{1}{12} \log _{\mathrm{e}}\left|\frac{\mathrm{t}-1}{\mathrm{t}+1}\right|+\mathrm{C}=\frac{1}{12} \log _{\mathrm{e}}\left|\frac{\mathrm{x}^{6}-1}{\mathrm{x}^{6}+1}\right|+\mathrm{C}\)
(B) : Given, \(I=\int \frac{x^{5} d x}{\left(x^{2}+x+1\right)\left(x^{6}+1\right)\left(x^{4}-x^{3}+x-1\right)}\) \(=\int \frac{x^{5} d x}{\left(x^{6}+1\right)\left(x^{3}+1\right)(x-1)\left(x^{2}+x+1\right)}=\int \frac{x^{5} d x}{\left(x^{6}+1\right)\left(x^{6}-1\right)}\) Let, \(x^{6}=t\) \(6 x^{5} d x=d t\) So, \(=\frac{1}{6} \int \frac{\mathrm{dt}}{(\mathrm{t}+1)(\mathrm{t}-1)}\) \(=\frac{1}{12} \log _{\mathrm{e}}\left|\frac{\mathrm{t}-1}{\mathrm{t}+1}\right|+\mathrm{C}=\frac{1}{12} \log _{\mathrm{e}}\left|\frac{\mathrm{x}^{6}-1}{\mathrm{x}^{6}+1}\right|+\mathrm{C}\)