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WolframAlpha can give step by step solution for indefinite integral. There seems to be similar question but for derivatives. Is there a way that I can generate my own step by step solution for indefinite integrals (for simple integrals having closed form)?

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If you have an answer for indefinite integral, just take its derivative and you will get the steps of the integration in backward direction. – Vahagn Poghosyan Sep 20 '13 at 20:45
@VahagnPoghosyan that's not what W|A gives ... does it? – Santosh Linkha Sep 20 '13 at 20:45
Sorry, but I have no Wolfram Alpha. It is not a free software and I can't find a cracked version. I gave you only theoretical idea. – Vahagn Poghosyan Sep 20 '13 at 20:48
@VahagnPoghosyan yes i thought of that too ... but if I post that as solution, that will definitely get rejected. I mean I just want to know if it is possible or not. I don't know coding in Mathematica and although I have heard that Risch algorithm is used to find closed solution of Indefinite integrals, so far I haven't encountered myself. If it's not easily done then ... i should give up. – Santosh Linkha Sep 20 '13 at 20:51
When I was a student, I used this trick of taking integrals during lectures and exams. I simply took integral by MATHEMATICA, then took its derivative step-by-step, and got steps of the integration :) I think it is possible to easily realize the code. – Vahagn Poghosyan Sep 20 '13 at 21:02

1 Answer 1

RUBI, the RUle-Based Integrator package, does what you want.

You can download a copy of the rules to use with Mathematica at this link: Follow the instructions to install and make sure that you select the option to "Show Steps". The package's Int command performs indefinite integration and will also show the substitution rules it used to get to the final result.

Suppose for instance that you want to obtain the antiderivative from $\int{x^2 \sin{x} \ \text{d} x}$. Using RUBI's Int command:

Int[x^2 Sin[x], x]

step 1

This has applied one of the rules in RUBI's rule base to simplify the integral. The intermediate results can be simplified further by evaluating them in turn:

-x^2 Cos[x] + Dist[2, Int[x Cos[x], x], x]

step 2

-x^2 Cos[x] + 2 x Sin[x] - Dist[2, Int[Sin[x], x], x]

step 3

Once further evaluations do not change the expression any more, the last output is the antiderivative we sought.

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