<s>[INST] Integral problem. Unsure of the approach. I have this integral:
$$\int_0^1 \frac{1 + 12t}{1 + 3t}dt$$
I can split this up into:
$$\int_0^1 \frac{1}{1+3t} dt + \int_0^1\frac{12t}{1+3t}dt$$
The left side:
$u = 1+3t$ and $du = 3dt$ and $\frac{du}{3} = dt$
so $$\frac{1}{3}  \int_0^1 \frac{1}{u} du = \frac{1}{3} \ln |u| + C$$
But what about the right?
 [/INST] My first attempt is to split the integral in another way:
$$\int_0^1{\frac{1+12t}{1+3t}}dt = \int_0^1{\frac{1+3t+9t}{1+3t}} dt = \int_0^1 1 \, dt + \int_0^1 \frac{9t}{1+3t} dt$$
Now observe that
$$\int_0^1 \frac{9t}{1+3t} dt = \int_0^1{\frac{3t}{1+3t}} \cdot 3dt = \int_0^1{\frac{3t}{1+3t}}d(3t) = \int_0^3 \frac{u}{1+u} du$$
, which can be found by another substitution, or
$$\int_0^3 \frac{u}{1+u} du = \int_0^3 \frac{u+1-1}{1+u} du = {\int_0^3 1\,du} - {\int_0^3{\frac{1}{1+u}} du} = 3-\int_1^4 \frac{1}{s} ds = 3-\log(4).$$
Hence $$\int_0^1 \frac{1+12t}{1+3t}dt = 1+3-\log(4)=4-\log(4).$$
</s>