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?


ANSWER

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


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
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