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Mathematics Wolfram|Alpha Wolfram Language

Problem solving an integral with Wolfram|Alpha?

Posted 8 years ago

Hi, today i was solving some math problems with integrals and wolfram alpha (windows 10 app version) gave me an answer that i think is incorrect, could this be possible? Here is the (i think wrong) result: http://s17.postimg.org/hg54ttf4u/Result.jpg After that i checked on another integral calculator and it gave me this result: http://s22.postimg.org/t1z74y041/Result_2.png which i think its correct. Can someone tell me if i am right or if i am ignoring something? Attachments:

POSTED BY: Rodrigo Viana

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Posted 8 years ago

Hi, that is the step by step solution. This also works: FullSimplify[1/2 Log[x + 2] - (Evaluate@Integrate[1/(2 x + 4), x]), Assumptions -> x > -1] It gives -(Log[2]/2) so a constant which obviously equals -(-(Log[2]/2) + Log[4]/2) You can also do: 1/2 Log[x + 2] + Log[2]/2 == (Evaluate@Integrate[1/(2 x + 4), x]) // FullSimplify or evaluate the derivatives: FullSimplify[D[1/2 Log[x + 2], x] == D[(Evaluate@Integrate[1/(2 x + 4), x]), x]] Cheers, M.

Hi,

that is the step by step solution. This also works:

FullSimplify[1/2 Log[x + 2] - (Evaluate@Integrate[1/(2 x + 4), x]), Assumptions -> x > -1]

It gives

-(Log[2]/2)

so a constant which obviously equals

-(-(Log[2]/2) + Log[4]/2)

You can also do:

1/2 Log[x + 2] + Log[2]/2 == (Evaluate@Integrate[1/(2 x + 4), x]) // FullSimplify

or evaluate the derivatives:

FullSimplify[D[1/2 Log[x + 2], x] == D[(Evaluate@Integrate[1/(2 x + 4), x]), x]]

Cheers,

POSTED BY: Marco Thiel

Posted 8 years ago

Dear Rodrigo, that's a kind of problem that comes up from now to then. It is true that the two results look different (and are up to a constant). If you plot the functions they do look different: Plot[{1/2 Log[x + 2], Evaluate@Integrate[1/(2 x + 4), x]}, {x, -5, 6}] They are obviously not identical. I will argue that they are still the same up to an additive constant. Let's look at WolframAlpha's solution the the solution that you got from the other program (and that I got from Mathematica using Integrate); I'll evaluate the difference for some arbitrary value, here $x=0$: (1/2 Log[x + 2] - (Evaluate@Integrate[1/(2 x + 4), x])) /. x -> 0 We get If we subtract that appropriately we get: Plot[1/2 (-2 (Log[2]/2 - Log[4]/2) + Log[2 + x] - Log[4 + 2 x]), {x, -15.5, 11.5}] That's what we can call a "numerical zero". We can now do that more rigorously: FullSimplify[1/2 Log[x + 2] - (Log[2]/2 - Log[4]/2) - (Evaluate@Integrate[1/(2 x + 4), x])] evaluates to zero. So the two solutions are identical up to an additive constant. Cheers, Marco