∫ x^2/(1 - 2x^3) dx = (-1/6) ln( 1-2x^3 ) + C
]]>But I didn't get around to solving "Student"s integral. Which I will attempt now ...
∫ x^2/(1 - 2x^3) dx
Try: u=1-2x^3
Then: du = -6x^2 dx (ie we differentiated it)
So: x^2 dx = (-1/6) du
Substituting 1-2x^3=u and x^2 dx = (-1/6) du we get:
∫ x^2/(1 - 2x^3) dx = ∫ 1/u (-1/6) du
Then it is fairly easy to solve ∫ 1/u (-1/6) du:
∫ 1/u (-1/6) du = (-1/6) ∫ 1/u du = (-1/6) ln u
Now put back u=1-2x^3:
(-1/6) ln( 1-2x^3 )
DONE! (I hope!)
]]>∫(x²-29x+5)/[(x-4)²(x²+3)]dx
this tuff question was on my calculus2 final, and no, i did not solve it
]]>∫ x^2/(1 - 2x^3) dx
is that right?
Otherwise it is too easy
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