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?

  • $\begingroup$ You could've written $1+12t$ as $4(1+3t)-3$ from the beginning. $\endgroup$ – stressed out Feb 22 '19 at 15:05

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.)

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:

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


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

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$$\frac{12t}{1+3t} = 4 - \frac{4}{1+3t}.$$ 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! ;)

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

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$$ \begin{align} \int\frac{1+12t}{1+3t}\,dt &=\int\frac{1}{1+3t}\,dt+12\int\frac{t}{1+3t}\,dt\\ &=\frac{1}{3}\int\frac{1}{1+3t}\frac{d}{dt}(1+3t)\,dt+\frac{12}{3}\int\frac{3t}{1+3t}\,dt\\ &=\frac{1}{3}\int\frac{1}{u}\,du+4\int\frac{1+3t-1}{1+3t}\,dt\\ &=\frac{1}{3}\ln{|u|}+4\left(\int\frac{1+3t}{1+3t}\,dt-\int\frac{1}{1+3t}\,dt\right)\\ &=\frac{1}{3}\ln{|u|}+4\int\,dt-\frac{4}{3}\int\frac{1}{1+3t}\frac{d}{dt}(1+3t)\,dt\\ &=\frac{1}{3}\ln{|u|}+4t-\frac{4}{3}\int\frac{1}{u}\,du\\ &=\frac{1}{3}\ln{|u|}+4t-\frac{4}{3}\ln{|u|}\\ &=\ln{\left(\frac{|u|^{1/3}}{|u|^{4/3}}\right)}+4t\\ &=\ln{|u|^{-1}}+4t\\ &=4t-\ln{|1+3t|}+C. \end{align} $$

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  • $\begingroup$ (Although the original is a definite integral.) $\endgroup$ – Teepeemm Feb 22 '19 at 15:04
  • 1
    $\begingroup$ As long as you know what the antiderivative is, applying the Newton-Leibniz formula is a trivial operation. $\endgroup$ – Michael Rybkin Feb 22 '19 at 15:06

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