Q.281.Indefinite Integration.

Question 281.
If k ∈ N and x > 0, then evaluate the following integral.
∫ [x^(3k) + x^(2k) + x^(k) ] * [ 2 x^(2k) + 3 x^(k) + 6 ]^(1/k) dx.

Answer 281.
∫ [x^(3k) + x^(2k) + x^(k) ] * [ 2 x^(2k) + 3 x^(k) + 6 ]^(1/k) dx
= ∫ [x^(3k - 1) + x^(2k - 1) + x^(k - 1) ]
                                   *[ 2 x^(3k) + 3 x^(2k) + 6x^(k) ]^(1/k) dx
[Taking x common from the first bracket and shifting in the second bracket as [x^(k)]^(1/k)]


Let 2 x^(3k) + 3 x^(2k) + 6x^(k) = u
=> 6k * x^(3k-1) + 6k * x^(2k-1) + 6k * x^(k-1) dx = du

=> Integral
= (1/6k) ∫ u^(1/k) du
= (1/6k) * [1/(1/k+1)] u^(1/k+1) + c
= [1/6(k+1)] * [2 x^(3k) + 3 x^(2k) + 6x^(k)]^[(k+1)/k] + c.
= [x^(k+1)/(6k+6)] * [2x^(2k) + 3x^(k) + 6]^(1/k + 1) + c.

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