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Evaluate the integral question inside!
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\[\frac{x^{2} dx }{x^3-e^3}\]
e^3 is a constant
Clue: u is usually thingy on the bottom u would have a du on top
Here is my answer: \frac{1}{3} ln (x^{3} - e^{3}) + C
u = x^3 - e^3 ; du = 3x^2 dx ; dx = du/3x^2
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eek!
x^2 du ------- 3x^2 u du ----- = (1/3) ln(u) +C 3 u
i wonder: (x^3-e^3) is a difference of cubes; can we split it to simplify further?
(x-e)(x^2 +x e +e^2) :)
yay! that's the answer I have! \[\frac{1}{3} \ln (x^{3} - e^{3} ) + C\]
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leave it :)
hmph! Not the final answer?
lol!
Thank you so much for the confirmation!
should we put the 1/3 as an exponent inside the ln?
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i dont see much more we can do to simplify it :)
Me either! Phew, you had me worried! lol
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