Question

If m any natural number, then the value of the integral$\int_{}^{}{(x^{3m} + x^{2m} + x^{m})}$ (2x^2m + 3x^m+ 6)^1/m dx is –

• A. $\frac{1}{6(m + 1)}${2x^3m + 3x^2m + 6x^m}^(1/m)+1 + C
• B. $\frac{1}{6m}${2x^3m + 3x^2m + 6x^m} ^(1/m)+1 + C
• C. $\frac{1}{6m}${2x^3m + 3x^2m + 6x^m}^1/m + C
• D. None of the above

Answer 1

Answer: (A)

$\frac{1}{6(m + 1)}${2x^3m + 3x^2m + 6x^m}^(1/m)+1 + C

Answer 2

$\int \left( x ^ { 3 m } + x ^ { 2 m } + x ^ { m } \right)$.dx

(2x^3m +3x^2m +6x^m)^1/m dx

$\frac { 1 } { 6 }$ $\int \left( 2 x ^ { 3 m } + 3 x ^ { 2 m } + 6 x ^ { m } \right) ^ { 1 / m }$(6x^3m–1 + 6x^2m–1 + 6x^m–1)dx

= $\frac { 1 } { 6 \mathrm {~m} }$ $\int \left( 2 x ^ { 3 m } + 3 x ^ { 2 m } + 6 x ^ { m } \right) ^ { 1 / m }$d (2x^3m + 3x^2m + 6x^m)dx– (2x^3m + 3x^2m + 6x^m)^1/m+1 + C