Evaluate the integral: $\int_{0}^3\frac{dx}{5x+1}$

$\int_{0}^3\frac{dx}{5x+1}$

So I cannot evaluate this integral because we do not have any rules for taking the anti-derivative of a quotient of functions. Thus, we need to separate this into two separate fractions before we can take the anti-derivative.

How do we do this?

$\frac{1}{5x+1}$ I can't transfer the binomial to the top unless I can do the following, and if so please make me aware that I am right.

$\frac{(5x+1)^{-1}}{1}$

Is this "illegal math"

• HINT:$$\int \frac{dx}{x}=\ln |x|$$ Commented Sep 20, 2015 at 15:45
• I would like to logically understand why the natural logarithm comes into play. Is it because we covert the integrand by U-substitution into a fraction the natural logarithm has a conversion factor for? Commented Sep 20, 2015 at 15:58
• It's because $ln(f(x)) = \frac{f'(x)}{f(x)}$, so $\int \frac{f'(x)}{f(x)}dx = \ln(x)$ Commented Sep 20, 2015 at 16:19

A minor point: It is unnecessary to write $\frac{(5x+1)^{-1}}{1}$ with a $1$ in the denominator. For any number $a$ it is always true that $a = \frac{a}{1}$ so we write simply $(5x+1)^{-1}$.

As for your main question, you probably know by know that you should make the substitution $5x+1 = t$. You are correct that there is no specific rule to integrate the quantity that you have. So what you want to do is pattern-match what you have to an integral form that you recognize. Ideally you recall that the derivative of $\ln(x)$ is $\frac{1}{x}$, and in that capacity you recognize the integral $\int \frac{1}{x} \text{d}x$. This is why the substitution $5x+1 = t$ is useful, because it converts your problem into one that you can easily anti-differentiate. Plugging in this substitution yields $$\int_0^3 \frac{1}{t} \text{d}x$$ but before continuing there are still two things to address. $(1)$ We need $\text{d}t$ not $\text{d}x$, as we are now trying to integrate with respect to $t$, and $(2)$ the integral limits from $0$ to $3$ were meant for your original quantity, not your substituted quantity. These will need to change to match your $t$ substitution.

Getting $\text{d}t$ is pretty easy. Take $5x+1 = t$ and differentiate both sides. You'll get $$5\text{d}x = 1\text{d}t \implies \text{d}x = \frac{1}{5}\text{d}t$$

For the new limits again take $5x+1 = t$. When $x = 0$ we have $5(0) +1 = t \implies t = 1$ and when $x = 3$ we have $5(3)+1 = t \implies t = 16$, so your new limits of integration are from $1$ to $16$. We now have everything we need to set up the new integral. \begin{align} \int_0^3 \frac{1}{5x+1}\text{d}x &\equiv \int_1^{16} \frac{1}{t}\left(\frac{1}{5}\text{d}t\right) \\ &= \frac{1}{5} \int_1^{16} \frac{1}{t}\text{d}t \end{align}

If this process is new to you, take note. It's (probably) the most common method you will use to solve integrals: Pattern-match your integral to one that you know how to anti-differentiate, make a substitution, solve for your infinitesimal in terms of the substitution, solve for the new limits of integration and lastly integrate.

• Great explanation. I now see the connection with using U-substitution so as the replicate the derivative of ln(x) = $\frac{1}{x}$. Nonetheless, my instructors have been using du. For example, u= 5x+1 and du (the derivative of 5x+1) = 5. Is there a reason why you have chosen to use dt instead? Commented Sep 20, 2015 at 16:23
• @Cetshwayo Yes the $u$ substitution is the standard mathematical convention. That is how I learned it as well. MSE seems to favor the $t$ substitution but there is no difference. $t$ is just a place-holder; it doesn't matter what we call it as long as we can pack the quantity $5x+1$ into it for the sake of integration. One correction I have to make about your comment above: The derivative of $u = 5x+1$ should be $\text{d}u = 5\text{d}x$ Commented Sep 20, 2015 at 16:25
• It may sound silly but I have a hard time comprehending the phrase "with respect to t". Can you say it another way so that I can better understand integration; indeed the Calculus concept. Thanks once again. Commented Sep 20, 2015 at 16:33
• I can try! "With respect to $t$" or "WRT $t$" simply means that you are going about your math problem treating $t$ as the variable. When you differentiate a function like $y = cx$ what you are being asked to do is "Differentiate the function $y$ with respect to $x$" which again means "treat $x$ as the variable". We'd have $y' = c$. However if I asked you to differentiate $y = cx$ with respect to $c$, we'd treat $c$ as the variable and $x$ as the constant, getting $y ' = x$. It's easy to lose track of this mindset, but it is important to always know which thing we are treating as the variable. Commented Sep 20, 2015 at 16:40

Notice, let $5x+1=u\implies 5dx=du$ $$dx=\frac{du}{5}$$

For lower limit $x=0\implies u=5(0)+1=1$

For upper limit $x=0\implies u=5(3)+1=16$

Now, substituting the corresponding values, we get $$\int_{0}^{3}\frac{dx}{5x+1}=\int_{1}^{16}\frac{1}{u}\frac{du}{5}$$ $$=\frac{1}{5}\int_{1}^{16}\frac{du}{u}$$ $$=\frac{1}{5}[\ln|u|]_{1}^{16}$$ $$=\frac{1}{5}[\ln|16|-\ln|1|]=\frac{1}{5}[\ln|16|-0]$$ $$=\color{}{\frac{1}{5}\ln(16)}=\color{}{\frac{1}{5}\ln(2^4)}$$ $$=\color{red}{\frac{4}{5}\ln 2}$$

• Why have your bounds increased on the evaluation bar? Commented Sep 20, 2015 at 16:06
• Alright, notice we assume $u=5x+1$ hence for lower bound $x=0\implies u=5(0)+1=1$. For upper bound, $x=3\implies u=3(5)+1=16$ Commented Sep 20, 2015 at 16:09
• Good to know. Thus the original integrand is used to re set the bounds by plugging the values of the bounds within the integral function and using the product of that as new bounds. However, Is there a reason for that? Commented Sep 20, 2015 at 16:15
• The only problem I have with this answer is that the explanation becomes condense at the latter part. Other then that, spot on. Commented Sep 20, 2015 at 16:55

Let $5x+1 = t\;,$ Then $d(5x+1) = dt\Rightarrow \displaystyle \frac{d}{dx}(5x+1)dx = dt$

So we get $$\displaystyle 5dx=dt\Rightarrow dx = \frac{1}{5}dt$$ and changing limit

We get $$\displaystyle \frac{1}{5}\int_{0}^{16}\frac{1}{t}dt = \frac{1}{5}\left[\ln |t|\right]_{1}^{16} = \frac{1}{5}\left[\ln|16|-\ln|1|\right] = \frac{1}{5}\ln(16)$$

No it isn't illegal to do that, it is the same thing. To solve sub $u=5x+1$ then $du=5dx$ which means $1/5du=dx$ Your integral becomes $\frac 1 5\int_{1}^{16}\frac{du}{u}$ and use the fact that $\int\frac {du} u=ln x$, should be absolute value of x in that last expression.

• Why have the bounds, [a,b] increased to [1,16] on the evaluation bar? Commented Sep 20, 2015 at 16:08
• @Cetshwayo at $x=0$, we get $u=1$ and at $x=3$, we get $u=16$. Since we have changed the variable from $x$ to $u$, we should write $u$'s limits and not that of $x$' s! Commented Sep 20, 2015 at 16:27

Hint:

with the substitution $5x+1=t$ you have $5dx=dt$ and the integral becomes: $$\int \dfrac{1}{5}\cdot \dfrac{dt}{t}$$

can you integrate?

• i believe you would want to change the limits of integration when you do your substitution. Commented Sep 20, 2015 at 15:55
• @MegaboofMD: Thanks. They was added in an edit. My first hint was witout limits. Commented Sep 20, 2015 at 15:58