Integrate` int x^3 sqrt (3x^2 + 2) dx`

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should come up with the substitution such that:

`sqrt (3x^2 + 2)  = t => 6x/(2sqrt(3x^2 + 2)) dx = dt`

`3x^2 + 2 = t^2 => 3x^2 = t^2 - 2=> x^2 = (t^2 - 2)/3`

`3x/(sqrt(3x^2 + 2)) dx = dt => 3xdx = tdt => xdx = (tdt)/3`

`int (x^3*sqrt (3x^2 + 2)) dx = int x^2*sqrt (3x^2 + 2)* xdx`

Changing the variable yields:

`int (t^2 - 2)/3*t*(tdt)/3 = (1/9) int t^2(t^2 - 2) dt`

`(1/9) int t^2(t^2 - 2) dt = (1/9) int (t^4 - 2t^2)dt`

`(1/9) int (t^4 - 2t^2) dt = (1/9) (int t^4 dt- int 2t^2 dt)`

`(1/9) int (t^4 - 2t^2) dt = (1/9) (t^5/5 - 2t^3/3 + c)`

Substituting `sqrt (3x^2 + 2)`  for t yields:

`int (x^3*sqrt (3x^2 + 2)) dx = (1/9) ((3x^2 + 2)^2sqrt(3x^2+2))/5 - 2((3x^2+2)sqrt(3x^2+2))/3 + c)`

Hence, evaluating the given integral using the substitution `sqrt (3x^2 + 2) = t`  yields `int (x^3*sqrt (3x^2 + 2)) dx = (1/9) ((3x^2 + 2)^2sqrt(3x^2+2))/5 - 2((3x^2+2)sqrt(3x^2+2))/3 + c).`

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