## Trending News

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

Integrate:

r^2

------- dr

(r+4)

### 6 Answers

- Anonymous1 decade agoFavorite 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!

- MechEng2030Lv 71 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).

- TheSicilianSageLv 71 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

- glassburnLv 44 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.

- How do you think about the answers? You can sign in to vote the answer.
- RyomaLv 71 decade ago
∫ [ x^2 / (x + 4) ] dx

= ∫ [ x - 4 + (16 / (x + 4)) ] dx

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