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Question
Integrate the following with respect to x :
`x^2/(1 + x^6)`
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Solution
`int x^2/(1 + x^6) "d"x = int x^2/(1 + (x^3)^2) "d"x`
Put x3 = u
3x2 dx = du
x2 dx = `1/3 "du"`
`int x^2/(1 + x^6) "d"x = int (1/3 "du")/(1 + "u"^2)`
= `1/3 int "du"/(1 + "u"^2)`
= `1/3 tan^-1 ("u") + "c"`
`int x^2/(1 + x^6) "d"x = 1/3 tan^-1 (x^3) + "c"`
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