Find the integral:
t^2 + 4 / t^3 +12t +8 dt
Do you know how an integral works?
They confuse me.
Is it \[\int\limits_{}^{}{t^2+4 \over t^3+12t+8} dt?\]
@AnwarAL it is what you posted. I think you can do u-substition for the integral.
OK let's call the integral I .. if you look closely you will see that The numerator is the derivative of the denominator after multiplying it (The numerator) by 3.. SO, multiply The numerator by 3 and divide outside the integral by 3, you get: \[I={1 \over 3}\int\limits_{}^{}{3t^2+12 \over t^3+12t+8}dt\] now you can see that The numerator is exactly the derivative of the numerator, then: \[I={1 \over 3} \ln \left| t^3+12t+8 \right|+c\]
you can do substitution u=t^3+12t+8 as smorlaz said.. the way I used looks faster and easier to me. They both lead to the same result.
does that make sense to you?
Wait a minute, why can't you just seperate the terms and integrate individually?
Or did I read the problem incorrectly?
^^ I think you did :)
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