∫(x7m+ x2m + xm).(2x6m + 7xm + 14)1/m.dx
(Have seen these sort of Qn(s) b4 too,,,A hint to start wud be enuf...)
1given integral can be written as \int (x^{7m}+x^{2m}+x^{m})\frac{(2x^{7m}+7x^{2m}+14x^{m})^{1/m}}{x}dx
= \int (x^{7m-1}+x^{2m-1}+x^{m-1}){(2x^{7m}+7x^{2m}+14x^{m})^{1/m}}dx
put t = (2x^{7m}+7x^{2m}+14x^{m})
now this can be done
29\int (x^{7m} + x^{2m} + x^{m})(2x^{6m}+7x^{m} +14)^{1/m} = \int x^{m}(x^{6m} + x^{m} + 1)(2x^{6m}+7x^{m} +14)^{1/m}
now write
\int x^{m}(x^{6m} + x^{m} + 1)(2x^{6m}+7x^{m} +14)^{1/m} = \int x^{m-1}(x^{6m} + x^{m} + 1)(2x^{7m}+7x^{2m} +14x^{m})^{1/m}
And then substitute xm = t
