# Integration of $\ln$ functions

Hi Math Stack Exchange,

I have a question about integration of $$\ln$$ functions. I am not sure how to integrate $$\ln$$ functions or the process involved behind it. My teacher said to simply insert the question into a graphics calculator to get an answer, however, I would like to know the step by step process to do this too by hand. If you could please assist me that would be great. Below is the question which I would like to Integrate.

$$\int_0^2 (18.87\ln(7.97x + 22) - 57.941) dx$$

The answer which I got after inserting the question into the Graphics Calculator is $$4157.59$$ if that helps at all.

Thank you so much!

• Have you learned integration by parts? – KM101 Feb 29 at 8:28
• @KM101 no I have not learned integration by parts at school. I have tried to self-learn it through Khan Academy but I don't understand it. I know how to integrate normal functions, so just x, but when Logs are associated I am lost... – THIS_GAMES_DOODOO Feb 29 at 8:38
• @KM101 I haven't learned integration by parts at all, can you please help me with it? – THIS_GAMES_DOODOO Feb 29 at 8:44
• You should check this. It gives quite a few examples and explains how integration by parts is really just the "inverse" of the product rule. – KM101 Feb 29 at 8:45

The integral is essentially $$\int_{0}^{2} ( a \, \ln(b x + c) + d) \, dx$$ and leads to: \begin{align} I &= a \, \int_{0}^{2} \ln(b x + c) \, dx + d \, \int_{0}^{2} dx \\ &= \frac{a}{b} \, [(b x + c) \, \ln(b x + c) - b x]_{0}^{2} + d \, [x]_{0}^{2} \\ &= \frac{a}{b} \, [(2 b + c) \, \ln(2 b + c) - c \, \ln(c)] + 2 (d - a) \\ &= 2 a \, \ln(2 b + c) + \frac{a c}{b} \, \ln\left(\frac{2 b + c}{c}\right) + 2(d - a). \end{align} From here it is just a matter of placing values into the formula for $$a, b, c, d$$.

To demonstrate the log integral value: differentiate the answer as follows \begin{align} \frac{d}{dx} \frac{1}{b} \, \left( (b x + c) \, \ln(b x + c) - b x \right) &= \frac{1}{b} \, \left( (b x + c) \, \frac{b}{b x + c} + b \, \ln(b x + c) - b \right) \\ &= \ln(b x + c). \end{align} Integrating both sides leads to the result presented.

Taking the result of the integral a few steps further leads to $$\int_{0}^{2} ( a \, \ln(b x + c) + d) \, dx = 2 \, a \, \ln\left((2 b + c)^p e^{-q} \right),$$ where \begin{align} p &= 1 + \frac{c}{2 b} \\ q &= \frac{c}{2 b} - \frac{d}{a} + 1. \end{align}

The product rule for differentiation says:

$$(uv)' = uv'+ u'v.$$

If you let $$u'=1$$ and $$v=\ln x$$ you have that $$u=x$$ and $$v'= 1/x$$ and get

$$(x\ln x)' = x\frac{1}{x} + 1\cdot \ln x.$$

Anti-differentiate this to get

$$x\ln x = \int 1 + \ln x \; dx$$

$$x\ln x = x + \int \ln x \; dx.$$

Now you know that

$$\int \ln x \; dx = x \ln x = x\ln x - x.$$

Now to integrate $$\ln (ax+b)$$ you substitute $$y=ax+b$$ to get

$$\int \ln(ax+b) \; dx = \frac{1}{a} \int \ln y \; dy = \frac{1}{a} (y\ln y -y)$$

$$\frac{1}{a}((ax+b) \ln(ax+b) -(ax+b)$$