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# help with finding the integer of x*((x^2)-2)

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hello, my problem is that when i try to do it i open it to x^3-2x and when i do integration it becomes x^4/4-x^2 which is wrong according to the answer given by the calculator which is (x^2-2)^2/4

help please! thanks!

Guest Oct 23, 2015

### Best Answer

#1
+26329
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If you expand (x^2-2)^2/4 you get x^4/4 - x^2 + 1

Although this looks different from x^4/4 - x^2 you have to remember that indefinite integrals are only defined to within a constant, so really, doing the integration your way you should have had x^2/4 - x^2 + k1 and the calculator should have had (x^2 - 2)^2/4 + k2 where k1 and k2 are (different) constants, related by k1 = k2+1.

Alan  Oct 23, 2015
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### 1+0 Answers

#1
+26329
+10
Best Answer

If you expand (x^2-2)^2/4 you get x^4/4 - x^2 + 1

Although this looks different from x^4/4 - x^2 you have to remember that indefinite integrals are only defined to within a constant, so really, doing the integration your way you should have had x^2/4 - x^2 + k1 and the calculator should have had (x^2 - 2)^2/4 + k2 where k1 and k2 are (different) constants, related by k1 = k2+1.

Alan  Oct 23, 2015

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