anonymous
  • anonymous
Find the anti derivative of (1 + 2x)^(2) I have got 2y^(3)/3 + 2y^(2) + y but wolfram alpha claims: https://www.wolframalpha.com/input/?i=integrate+%281%2B2x%29^%282%29 Where am i going wrong? I foiled it out to 4x^(2) + 4x + 1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
then took the anti derivative but I have no idea what I did wrong
anonymous
  • anonymous
also Im aware I didnt add plus c but it isn't important
experimentX
  • experimentX
i guess you are worng ... there is no y, and with what did you integrate the right answer would be https://www.wolframalpha.com/input/?i=integrate+%281%2B2x%29%5E%282%29+dx

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anonymous
  • anonymous
yeah pretend x = y
experimentX
  • experimentX
first of all .. expand that square and try to integrate each term with respect to x
ash2326
  • ash2326
We have \[\int (1+2x)^2 dx\] Let \[1+2x=t\] then \[2dx=dt\] or \[dx=\frac{dt}{2}\] so we have now \[\int (t)^2 \frac{dt}{2}\] We know that \[\int x^n=\frac{x^{n+1}}{n+1}\] so we get \[\int (t)^2 \frac{dt}{2}=\frac{t^3}{3 \times 2}+c\] we know t=1+2x so we get \[\int (1+2x)^2 dx=\frac{(1+2x)^3}{6}+c\]
anonymous
  • anonymous
oh thanks ash
anonymous
  • anonymous
wait why isn't it 2dt = dx?
ash2326
  • ash2326
We have assumed \[1+2x=t\] so differentiate both sides we get \[2dx=dt\] or \[dx=\frac{dt}{2}\]
anonymous
  • anonymous
I still dont get it :l
ash2326
  • ash2326
Okay We assumed \[1+2x=t\] Let's differentiate this with respect to x or \[\frac{d}{dx}(1+2x)=\frac{d}{dx}(t)\] We get \[\frac{d}{dx}(1)+\frac{d}{dx}(2x)=\frac{d}{dx}(t)\] We get now \[0+2=\frac{dt}{dx}\] Multiply both sides by dx \[2dx=dt\]
ash2326
  • ash2326
Does it help?
anonymous
  • anonymous
ok I think I see and yes a lot

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