Mathematics

# For any natural number m, the value of $\displaystyle\int \left(x^{3m}+x^{2m}+x^m\right)\left(2x^{2m}+3x^m+6\right)^{\frac{1}{m}}dx, x > 0$ is?

$\dfrac{1}{6(m+1)\cdot }\left(2x^{3m}+3x^{2m}+6x^m\right)^{\frac{m+1}{m}}+C$

##### SOLUTION
We are given,
$I=\displaystyle\int \left(x^{3m}+x^{2m}+x^m\right)\left(2x^{2m}+3x^m+6\right)^{\dfrac{1}{m}}dx$, $x > 0, m > 0$
$I=\displaystyle\int x(x^{3m-1}+x^{2m-1}+x^{m-1})\cdot (2x^{2m}+3x^m+6)^{1/m}dx$
(Taking x as common)
$I=\displaystyle\int (x^{3m-1}+x^{2m-1}+x^{m-1})\cdot (2x^{3m}+3x^{2m}+6x^m)^{1/m}dx$
(product of x with another function $2x^{3m}+3x^{2m}+6x^m)^{1/m}$)
$I=\displaystyle\int (x^{3m-1}+x^{2m-1}+x^{m-1})\cdot (2x^{3m}+3x^{2m}+6x^m)^{1/m}dx$
Let $2x^{3m}+3x^{2m}+6x^m=t$ (let differentiate)
$6m x^{3m-1}+6m x^{2m-1}+6m x^{m-1}dx=dt$
$6m(x^{3m-1}+x^{2m-1}+x^{m-1})dx=dt$
$\therefore (x^{3m-1}+x^{2m-1}+x^{m-1})dx=\dfrac{dt}{6m}$
$I=\displaystyle\int \dfrac{t^{1/m}}{6m}dt=\dfrac{t^{1/m+1}+c}{6m\left(\dfrac{1}{m}+1\right)}=\dfrac{t^{m+1/m}}{6+6m}+C$
So, $I=\dfrac{(2m^{3m}+3x^{2m}+6x^m)^{\dfrac{m+1}{m}}}{6(m+1)}+C$.

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
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