Integration by parts explanation

Question

I have he following question:

How did they get v prime to equal $x^n$ and they change v to be the integral of $x^n$

Answer 1

When you do integration by parts, you only choose one thing: $u$

After that, everything else is automatically determined. When you choose $u$, then $v'$ is everything that's left over in the integrand.

For thoroughness I should mention that some sources teach that $v'$ should be picked as "the most readily integrable piece of the integrand" and therefore $u$ is automatically determined. But I hate that approach because it's vague and there's actually a legitimate and very clear process for choosing $u$.

The LIATE rule is helpful for choosing $u$. Applying that rule tells us to choose $u = \ln x$ for the given integral.

So: $\displaystyle \int x^n \underbrace{\ln x}_u \, dx$

This means that $v'$ must be everything else remaining in the integrand: $\displaystyle \int \underbrace{x^n}_{v'} \underbrace{\ln x}_u \, dx$

Then to find $v$, you just integrate $v'$ using the power rule for integrals. (By convention we ignore the arbitrary constant of integration when going from $v'$ to $v$ in integration by parts because it will always cancel out anyway.)

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Edit: Here's the official journal article about LIATE. As of the time I post this you can read it for free if you register for a free JSTOR account.

Answer 2

The so-called integration by parts formula comes from the chain rule of differentiation: $$ (uv)' = u'v + uv'. $$ By the definition of primitive we can write the above as $$ uv + \text{constant} = \int u'v + \int uv', $$ equivalent to $$ \int uv' = uv - \int u'v + \text{constant}. $$ Now to apply this to a particular example like the present one $\int x^{n}\ln x$, it takes an observation to find out that taking $\ln x$ as $u$ and $x^{n}$ as $v$ is more convenient because $(\ln x)' = 1/x$ and $(\frac{x^{n+1}}{n+1})' = x^{n}$ directly.

Answer 3

They chose $\;u=\log x\;,\;\;v'=x^n\;$ , and then applied exactly the formula

$$\int x^n\log x\,dx=\overbrace{\frac{x^{n+1}\log}{n+1}}^{=uv}-\overbrace{\int\frac{x^{n+1}}{x(n+1)}dx}^{=\int u'v}=$$$${}$$

$$=\frac1{n+1}\left[x^{n+1}\log x-\int x^n\,dx\right]=\frac1{n+1}\left[x^{n+1}\log x-\frac{x^{n+1}}{n+1}\right]+C=$$

$$=\frac{x^{n+1}}{n+1}\left[\log x-\frac1{n+1}\right]+C$$

Answer 4

By comparing original integration with :

$$\int uv' ~~\text{and}~~ \int x^n \ln x$$

We get $v' = x^n , u = \ln x $

Now, $v'$ denotes derivative of $v$ ,

$$\frac{d{v}}{dx} = x^n \implies v = \int x^n dx$$

There is a priority rule for convenience of selecting $u$ :

$\text{ILATE}$ : Inverse Trig. > Logarithmic > Algebraic >Trigonometric Exponential >