3u integration (1 Viewer)

[want2beSMART]

hey how u integrate x(x^2 - 3 )^4

 

[FinalFantasy]

integrate x(x^2 - 3 )^4 dx
let u=x^2-3
du=2x dx
integrate x(x^2 - 3 )^4 dx=1\2 int. u^4 du........

 

[want2beSMART]

cool

haha well at first i expanded then simplified...and i got a really long answer haha

 

[withoutaface]

/
|x⁹-12x⁷+54x⁵-108x³+81x
/
x¹⁰/10-3x⁸/2+9x⁶-27x⁴+81x²/2

Yes, I just had to try it, and it's not that hard if you're good at arithmetic (I'm not, for anyone who's gonna point out an error in my solution).

 

[Slidey]

It's also interesting to compare both solutions. Justin's solution lacks a constant, while the substitution method has one. Yet both methods are valid.

How can this be so? Well, this is the reason you always attach a + C to your indefinite integrals. This unknown constant accounts for such variations, since C' (derivative of C, a constant) is 0.

 

[illin]

are they both valid?

integral=2/5*(x^2-3)^5 + C

 

[withoutaface]

It's also interesting to compare both solutions. Justin's solution lacks a constant, while the substitution method has one. Yet both methods are valid.

How can this be so? Well, this is the reason you always attach a + C to your indefinite integrals. This unknown constant accounts for such variations, since C' (derivative of C, a constant) is 0.

You know it was a clerical error, shush

 

[JamiL]

yer both r vaild but y would u waste ur tym... neway u will no because in 3u they will give u a hint. ie the question would be intergreate f(x)=x(x^2 -3)^4 using the sunbtitution u=x^2 - 3.... in other words they will give u u then it becomes realy easy. 4u u have 2 find it out by urself