Solveeit Logo
Login

Question

Question: For any natural number m \(\int_{}^{}{(x^{3m} + x^{2m} + x^{m})(2x^{2m} + 3x^{m} + 6})^{1/m}dx,(x >...

For any natural number m

∫(x3m+x2m+xm)(2x2m+3xm+6)1/mdx,(x>0)\int_{}^{}{(x^{3m} + x^{2m} + x^{m})(2x^{2m} + 3x^{m} + 6})^{1/m}dx,(x > 0) =

A

(2x3m+3x2m+6xm)m+1m6(m+1)\frac{(2x^{3m} + 3x^{2m} + 6x^{m})^{\frac{m + 1}{m}}}{6(m + 1)}

B

(2x3m+3x2m+6xm)mm+16(m+1)\frac{(2x^{3m} + 3x^{2m} + 6x^{m})^{\frac{m}{m + 1}}}{6(m + 1)}

C

−(2x3m+3x2m+6xm)m+1m6(m+1)- \frac{(2x^{3m} + 3x^{2m} + 6x^{m})^{\frac{m + 1}{m}}}{6(m + 1)}

D

−(2x3m+3x2m+6xm)mm+16(m+1)\frac{- (2x^{3m} + 3x^{2m} + 6x^{m})^{\frac{m}{m + 1}}}{6(m + 1)}

Answer

(2x3m+3x2m+6xm)m+1m6(m+1)\frac{(2x^{3m} + 3x^{2m} + 6x^{m})^{\frac{m + 1}{m}}}{6(m + 1)}

Explanation

Solution

I=∫(x3m−1+x2m−1+xm−1)(2x3m+3x2m+6xm)1/mdxI = \int_{}^{}{(x^{3m - 1} + x^{2m - 1} + x^{m - 1})(2x^{3m} + 3x^{2m} + 6x^{m})^{1/m}dx}Put 2x3m+3x2m+6xm=t⇒6m(x3m−1+x2m−1+xm−1)dx=dtI=∫16mt1/mdt=16mt1m+11m+1+c=16(m+1)t(m+1)/m+c2x^{3m} + 3x^{2m} + 6x^{m} = t \Rightarrow 6m(x^{3m - 1} + x^{2m - 1} + x^{m - 1})dx = dtI = \int_{}^{}{\frac{1}{6m}t^{1/m}dt = \frac{1}{6m}\frac{t^{\frac{1}{m} + 1}}{\frac{1}{m} + 1}} + c = \frac{1}{6(m + 1)}t^{(m + 1)/m} + cAgain put the value of t, then we get option (1).