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Had taken a break from studying, and to get back into it I did some review questions. Was evaluating this with integration by parts.

$$\int_1^3 (x-1)\ln x dx$$

Initially I chose $u = (x-1)$ and $\frac{dv}{dx} = \ln x$ as it led to a simpler expression.

However that is wrong and of course I should have chosen $u = \ln x$

But I can't see why the first choice was wrong; the algebra seemed to follow just fine but the answer was wrong.

I had the following expression following my initial incorrect choice of $u$ and $v$:

$$\left. (x-1)\frac{1}{x} \right\vert_1^3 - \int_1^3 \frac {1}{x} dx $$

Can anyone help me see where I've gone wrong? Or is it simply that the choice of $u$ and $v$ are simply wrong because the answer is wrong?

I've now refreshed the order of preference for choosing $u$ but this seems like an easy trap to fall into; in all previous problems I've done the incorrect choice was obvious as it led to something much more complicated or impossible (at this level).

Perhaps I've made a mistake elsewhere and welcome your input.

Thanks in advance.

  • 1
    If $\frac{dv}{dx}=\ln(x)$, why $v(x)=\frac{1}{x}$ ? What is $\int\ln(x)\,dx$ ? – 2017-02-13
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    The mistake you made is that $\frac{d}{dx}\ln{x} = \frac{1}{x}$, not $\int{\ln{x}}=\frac{1}{x}$! In your case $v$ would actually be $x\ln{x} - x$ – 2017-02-13
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    Oops, that was silly. Thank you both. – 2017-02-13

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Well, integration by parts:

$$\int\text{f}\left(x\right)\text{g}'\left(x\right)\space\text{d}x=\text{f}\left(x\right)\text{g}\left(x\right)-\int\text{f}\space '\left(x\right)\text{g}\left(x\right)\space\text{d}x\tag1$$

So, when we use $(1)$ in your question:

  1. When $\text{f}\left(x\right)=x-1$ and $\text{g}'\left(x\right)=\ln\left(x\right)$ we get: $\text{f}\space'\left(x\right)=1$ and $\text{g}\left(x\right)=x\ln\left(x\right)-x$
  2. When $\text{f}\left(x\right)=\ln\left(x\right)$ and $\text{g}'\left(x\right)=x-1$ we get: $\text{f}\space'\left(x\right)=\frac{1}{x}$ and $\text{g}\left(x\right)=\frac{x^2}{2}-x$

So, for the integral we get:

  1. $$\int_1^3\left(x-1\right)\ln\left(x\right)\space\text{d}x=x\left(x-1\right)\left(\ln\left(x\right)-1\right)\Big{|}_1^3-\int_1^3x\left(\ln\left(x\right)-1\right)\space\text{d}x=$$ $$6\left(\ln\left(3\right)-1\right)-\int_1^3x\left(\ln\left(x\right)-1\right)\space\text{d}x\tag2$$
  2. $$\int_1^3\left(x-1\right)\ln\left(x\right)\space\text{d}x=\ln\left(x\right)\left(\frac{x^2}{2}-x\right)\Big{|}_1^3-\int_1^3\frac{\frac{x^2}{2}-x}{x}\space\text{d}x=$$ $$\frac{3\ln\left(3\right)}{2}-\frac{1}{2}\int_1^3\left(x-2\right)\space\text{d}x\tag3$$