Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

integrate r^2 / (r+4) dr?

Integrate:

r^2

------- dr

(r+4)

6 Answers

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  • Anonymous
    1 decade ago
    Favorite Answer

    Note that:

    r^2/(r + 4)

    => [(r^2 - 16) + 16]/(r + 4)

    = (r^2 - 16)/(r + 4) + 16/(r + 4)

    = [(r + 4)(r - 4)]/(r + 4) + 16/(r + 4), by difference of two squares

    = (r - 4) + 16/(r + 4).

    Thus, the integral becomes:

    ∫ r^2/(r + 4) dr

    = ∫ [(r - 4) + 16/(r + 4)] dr

    = ∫ r dr - 4 ∫ dr + 16 ∫ 1/(r + 4) dr

    = (1/2)r^2 - 4r + 16ln|r + 4| + C.

    I hope this helps!

  • 1 decade ago

    ∫(r²)/(r + 4) dr = ∫(r² - 16 + 16)/(r + 4) dr = ∫(r - 4 + 16/(r + 4)) dr = r²/2 - 4r + 16*ln|r + 4| + C

    Note: Subtracting 16 and adding 16 is not changing the integral in any way or manner since (r² - 16 + 16) is just r².

    Also, notice that ∫(r² - 16 + 16)/(r + 4) dr can be split up into

    ∫(r² - 16)/(r + 4) dr + ∫16/(r + 4) dr

    and that's how I got my above answer.

    Keep in mind that r²-16 = (r + 4)(r - 4).

  • 1 decade ago

    let u = r+4; du = dr

    .. ∫ [ r² / (r+4) ] dr

    = ∫ [ (u-4)² / u ] du

    = ∫ [ (u² - 8u + 16) / u ] du

    = ∫ u du - 8 ∫ du + 16 ∫ 1/u du

    = ½ u² - 8u + 16 ln(u) + C

    = ½ (r+4)² - 8(r+4) + 16 ln(r+4) + C

  • 4 years ago

    hint: First is the unit circle and 2nd is lower back a circle r^2 = 2rsinT or x^2 +y^2 = 2y ability a circle with radius a million and middle (0,a million) yet once you decide on basically polars for use this would possibly not paintings.

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  • Ryoma
    Lv 7
    1 decade ago

    ∫ [ x^2 / (x + 4) ] dx

    = ∫ [ x - 4 + (16 / (x + 4)) ] dx

    = (x^2 / 2) - 4x + 16*ln|x + 4| + C

  • 1 decade ago

    holy crap my brain hurst after lookin at that

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