Integral problem. Unsure of the approach.

Hint:

$$\frac{12t}{1+3t} = 4 - \frac{4}{1+3t}.$$ Now use the same approach as you did for the first part. In fact you should’ve simplified the fraction before splitting into parts.

Remember to evaluate the integral at t=0 and t=1! ;)

At least on an initial glance, I'd make the substitution

$$u = 1+3t \implies du = 3dt \implies dt = \frac{du}{3}$$

Note that $t = (u-1)/3$ and thus $12t = 4(u-1)$.

Then the integral becomes

$$\int_0^1 \frac{12t}{1 + 3t} dt = \int_\ast^\ast \frac{4(u-1)}{u} \frac{dt}{3} = \frac 4 3 \int_\ast^\ast \frac{u-1}{u}du = \frac 4 3 \int_\ast^\ast 1 - \frac{1}{u}du$$

But this seems a bit excessive and unnecessary, even if it's the first thing to come to mind. (That's why I left the bounds as $\ast$'s: in cases like these, overcomplication suggests an alternative approach.)

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Another trick is to "add and subtract the same thing from the top"; this can be handy in a lot of math classes for manipulations if you add and subtract the right thing. In this case, we notice:

$$12t = 12t + 4 - 4 = 4(3t+1) - 4$$

Thus,

$$\int_0^1 \frac{12t}{1 + 3t} dt = \int_0^1 \frac{4(3t+1) - 4}{1 + 3t} dt = \int_0^1 4 - \frac{4}{1 + 3t} dt$$