Integral Calculus> integrate [{(x^3m) + (x^2m) + (x^m)} {2(x...

integrate [{(x^3m) + (x^2m) + (x^m)} {2(x^2m) + 3(x^m) + 6}^(1/m)] dx , x>0

vaanya , 11 Years ago

Grade 12

1 Answers

Jitender Singh

Ans:

$I = \int (x^{3m}+x^{2m}+x^{m})(2x^{2m}+3x^{m}+6)^{1/m}dx$

$I = \int (x^{3m}+x^{2m}+x^{m})((2x^{2m}+3x^{m}+6).\frac{x^{m}}{x^{m}})^{1/m}dx$

$I = \int (x^{3m-1}+x^{2m-1}+x^{m-1})(2x^{3m}+3x^{2m}+6x^{m})^{1/m}dx$

$(2x^{3m}+3x^{2m}+6x^{m})=t$

$(6mx^{3m-1}+6mx^{2m-1}+6mx^{m-1})dx=dt$

$(x^{3m-1}+x^{2m-1}+x^{m-1})dx=\frac{dt}{6m}$

$I = \frac{1}{6m}\int t^{1/m}dt=\frac{t^{\frac{1}{m}+1}}{6(m+1)}+c$

$I = \frac{(2x^{3m}+3x^{2m}+6x^{m})^{\frac{1}{m}+1}}{6(m+1)}+c$

Thanks & Regards

Jitender Singh

IIT Delhi

askIITians Faculty

Last Activity: 11 Years ago

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Integral Calculus> integrate [{(x^3m) + (x^2m) + (x^m)} {2(x...

integrate [{(x^3m) + (x^2m) + (x^m)} {2(x^2m) + 3(x^m) + 6}^(1/m)] dx , x>0

vaanya , 11 Years ago

Jitender Singh

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