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$$ \int{u\frac{dv}{dx}} \equiv uv -\int{v\frac{du}{dx}}$$ I prefer to write as: $ \int{uv} \equiv u\int{v} -\int{(\int{v})\frac{du}{dx}}$

I get how this works, $\int{x\ln{x} = \ln{x}\cdot\frac{x^2}{2} -\frac{x^2}{4}}$

However, when limits are applied, I don't seem to be able to get the right answer, doing:

$$\int_1^2{x\ln{x}} = \ln{x}\bigg[\frac{x^2}{2}\bigg]^2_1-\int^2_1{\bigg(\bigg[\frac{x^2}{2}\bigg]^2_1\frac{1}{x})} = \\ \int_1^2{x\ln{x}} = \bigg[\frac{\ln{x}\cdot x^2}{2}\bigg]^2_1-\int^2_1{\bigg(\bigg[\frac{x}{2}\bigg]^2_1\bigg)} = \\ \ln{4} - \cdots$$

How is the second part resolved? $- \int^2_1{\frac12}$??

$= - [\frac x2]^2_1 = -\frac12; \ln{4} - \frac12$ isn't the right answer.

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  • $\begingroup$ What does that mean?... $\endgroup$
    – Tobi
    Jan 4 '17 at 2:06
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There are two related but different operations you have to do for integration by parts when it's between limits: finding an antiderivative for one of the functions (and the derivative of the other, but that's not where your problem lies), and finding the value of the integral between the limits, which we normally (but not necessarily) do by finding a continuous antiderivative and plugging in the limits.

Remember that integration by parts comes from the product rule: $$ uv + C = \int (uv)' = \int u'v+\int uv', $$ where $u'$ means $du/dx$ and so on, and we conventionally don't worry about the integration constant until later. If we now put limits on this, it becomes $$ [uv]_a^b = \int_a^b u'v + \int_a^b uv' $$

Therefore to evaluate a definite integral $ \int_a^b fg $ using integration by parts, we need a function $F$ so that $F'=f$, i.e. an antiderivative of $f$, from which we find, using the previous displayed equation, that $$ \int_a^b fg = \int_a^b F'g = [Fg]_a^b - \int_a^b Fg'. $$ In particular, finding $F$ is the same as doing an indefinite integral, which you would write as $\int f$, and does not involve putting the limits in.

The moral of the story is that the computation you want is done as follows: $$\begin{align} \int_1^2 x\log{x} \, dx &= \left[\frac{1}{2}x^2\log{x} \right]_1^2 - \int_1^2 \frac{1}{2}x^2 \frac{1}{x} \, dx \\ &= \frac{2^2\log{2}}{2}-\frac{1^1\log{1}}{2} - \frac{1}{2}\int_1^2 x \, dx \\ &= 2\log{2}-0 - \left[ \frac{1}{4} x^2 \right]_1^2 \\ &= \log{4} - \frac{3}{4}. \end{align}$$

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If you prefer to write your way, it's fine as long as you keep in mind that, you're supposed to plug in the upper and lower limits right at the very end, once you're done with evaluating the integrals.

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Let's try it the way other people do it:

$$ \int u\frac{dv}{dx}\,dx = uv -\int v\frac{du}{dx}\,dx, $$ letting $u=\ln x$ and $v = \frac{x^2}2,$ gives us \begin{align} \int (\ln x)(x)\,dx &= (\ln x)\left(\frac{x^2}2\right) - \int \left(\frac{x^2}2\right)\left(\frac 1x\right)\,dx \\ &= \frac{x^2 \ln x}2 - \int \frac x2\,dx \end{align} Integrated over the interval $x\in[1,2]$ this becomes \begin{align} \int_1^2 x \ln x\,dx &= \left[\frac{x^2 \ln x}2\right]_1^2 - \int_1^2 \frac x2\,dx \\ &= \left(\frac{2^2 \ln 2}2 - \frac{1^2 \ln 1}2\right) - \left(\frac{2^2}4 - \frac{1^2}4\right) \\ &= (2 \ln 2 - 0) - \left(1 - \frac14\right) \\ &= \ln 4 - \frac 34. \end{align}

Your way is to write $\int v$ where other people write $v,$ so you write $v$ where they write $\frac{dv}{dx}.$ So far so good, but remember that in the two places where you write $\int v$ on the right-hand side, you must substitute the same antiderivative of $v$ in each place--that is, in the rightmost integral you must multiply $\frac{du}{dx}$ by the same function of $x$ that you multiply with $u$ in the first term on the right. You do not multiply $\frac{du}{dx}$ by a definite integral with given bounds, because that is a constant, not the desired function of $x.$

In short, when you make your substitutions for $u$ and $v$ the results should be just like the ones above.


While it was a mistake for you to write $\left[\frac{x^2}2\right]_1^2$ where you should have written just $\frac{x^2}2$ in the integral on the right-hand side, you made a second serious error on top of that: $$ \left[\frac{x^2}2\right]_1^2 \frac1x = \left(2 - \frac12\right) \cdot \frac1x = \frac3{2x}, $$ but $$ \left[\frac x2\right]_1^2 = 1 - \frac12 = \frac12. $$ The expression $\left[\frac x2\right]_1^2$ is not a legitimate substitute for $\left[\frac{x^2}2\right]_1^2 \frac1x$ unless you happen to be evaluating both in an environment where $x$ is a constant $3$ outside the square brackets. Of course you only had the opportunity to make this mistake because you put "integration bounds" in a place where they had no business to be in the first place, but you should understand why these two expressions are so fundamentally different.

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