how do you integrate - t ^ 2 / (1 + t ^ 3) ?!?!?

Question

well the negative can be moved to the outside of the integral but i still dont know what to do... i see that by parts the denominator could factor to (x + 1) (x^2 - x + 1) ... but then what!?!?

Answer 1

Integral (-t^2 / (1 + t^3) ) dt

First, let's move the negative outside of the integral. I'll represent it as -1, since that's what it effectively is.

(-1) * Integral ( [t^2 / (1 + t^3)] dt)

Now, I'm going to move the t^2 next to the dt. I'm doing this to show you how substitution is going to work.

(-1) * Integral ( [1/(1 + t^3)] t^2 dt )

Here is where we use substitution.
Let u = 1 + t^3. Then
du = 3t^2 dt. Multiplying both sides by (1/3), we have

(1/3) du = t^2 dt

Look how the tail end of our (current) integral is t^2 dt. As per the last step in our substitution, we're going to replace that with
(1/3) du.

Therefore, (-1) * Integral ( [1/(1 + t^3)] t^2 dt ) becomes

(-1) * Integral ( [1/u] (1/3) du)

Pulling out the constant (1/3) will merge it with the (-1), and we will get

(-1/3) * Integral ( [1/u] du)

This is now trivial to solve.

(-1/3) * ln|u| + C

Replacing u = 1 + t^3 back,

(-1/3) * ln|1 + t^3| + C

The | | bars mean "absolute value", since the integral of 1/x is equal to ln|x| + C.

Answer 2

u+1+t^3

Answer 3

you have to use u-substitution.
set u = 1+t^3, then du=3t^2dt
that makes the formula:
(-1/3) integral du/u
= (-1/3) ln u + C
then substitute u back in
=(-1/3) ln(1+t^3) + C
=ln(t+t^3)^(-1/3) + C

Answer 4

u = 1+ t^3
du/dt = 3t^2

du = 3t^2 dt

-t^2 = -du/3
1+t^3 = u

You can now write the equation as follows:

-du/3u

When u integrate the above u get,

-1/3 ln u + x

-1/3 ln (1+t^3) + c

Answer 5

- t ^ 2 / (1 + t ^ 3) = -(1/3) 3t^2/(1+t^3).
Its integral = - (1/3) ln(1+t^3) + Constant