anonymous
  • anonymous
Indefinite Integral Calculation: S=integral S(X^2+4x)/(x^3+6x^2+5)dx So I get that I have to do the u substitution: u = x^3 + 6x^2 +5 --> du = 3x^2 +12x --> 1/3du = x^2 +4x dx then --> 1/3 S 1/u du --> 1/3ln(u) --> then i plug u back in. The problem I am having is the symbolab calculator that I'm using to check my work is adding this extra step that I don't understand. it gets u = x^3 +6x^2 +5 then it gets du = 3x(x+4) then dx = 1/3x(x+4) du it plugs it in as: S (x^2+4x)/u (times) 1/(3x(x+4)du then 1/3u du, after which it lines up with what I got. help pl0x :)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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baru
  • baru
u get\(\int \frac{1}{3u}du\) ?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
I just don't understand why they solved for dx after getting du, cant you just plug du back in to get what you need?

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baru
  • baru
i really don't understand what is confusing, both methods are equal, the calculator has just added one extra step,
anonymous
  • anonymous
@baru I don't understand the one step thought, that's what I want to understand. @robtobey good one lol. Doesn't answer my question though
baru
  • baru
\(u=x^3+6x^2+5\\du=(3x^2+12x)dx\\du=3(x^2+4x)dx\\du=3x(x+4)dx\\dx=\frac{du}{3x(x+4)} \)
baru
  • baru
\(\int \frac{x^2+4x}{x^3+6x^2=5} dx\\\\rearrange~numerator\\\int \frac{x(x+4)}{x^3+6x^2=5} dx\)
baru
  • baru
substitute for for u and dx \(\int \frac{x(x+4)}{u} \frac{du}{3x(x+4)}\)
baru
  • baru
x(x+4) in the numerator and denominator get cancelled leaving \(\int \frac{du}{u\times 3}\)
anonymous
  • anonymous
@baru, it's true what they say.. you really are a human calculator. lol thanks

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