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Can I define the '$u = \dots$' for an integral?

For example when I integrate $\frac{x^3}{\sqrt{6 - x^2}}$ with respect to $x$, the program automatically sets $u = x^2$, how can I change this so it uses $u = (6 - x^2)$ or some other value?

Glorfindel
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SchpinkelDinkel
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    Related: https://mathematica.stackexchange.com/questions/146458/problem-with-evaluation-of-the-integral/146491#146491 There's also DChange, but I think you need to write the integral as a differential equation. – Michael E2 May 13 '18 at 16:34
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    I'm voting to close this question as off-topic because the question seems about how to formulate W|A input and not about how to use Mathematica to communicate with W|A. – Michael E2 May 13 '18 at 16:58

1 Answers1

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It is not clear what you are trying to do

expr = x^3/Sqrt[6 - x^2];

int = Integrate[expr, x]

(* -(1/3) Sqrt[6 - x^2] (12 + x^2) *)

Checking that int is an anti-derivative of expr

expr == D[int, x] // Simplify

(* True *)

The change of variables u == (6 - x^2) does not change the result

Integrate[
   (x^3/Sqrt[6 - x^2] /. x -> Sqrt[6 - u])* D[Sqrt[6 - u], u],
   u] /. u -> 6 - x^2 // Simplify

(* -(1/3) Sqrt[6 - x^2] (12 + x^2) *)

You can add any arbitrary constant to int and still have a valid anti-derivative

Bob Hanlon
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  • Clearly, the OP does not use Mathematica. What the OP is trying to ask is how to perform the change of variables using W|A. I'm having this issue too – Cheng Feb 09 '21 at 02:57