integral

[spacewater]

X^4 ln x dx
u = x^4 du = 4x^3 dx
v = 1/x dv = ln x dx
x^4(1/x)-4 (x^3/x) dx
x^3-4(1/4x^3)+c
x^3-4/3x^3+c = x^3(1-(4/3)+C

or

u = ln x du = 1/x dx
v = x^5/5 dv = x^4 dx
ln x (x^5/5) - 1/5 (x^5/x) dx
1/5(x^5ln x) - 1/25x^5 + C

If there an easier way to tell which one is "u" and "uv" ? The first one is the one i actually did and the second one is the correct one. Can someone please point out how do find u and dv a bit faster than solving the entire problem? Thanks

 

[BigGlenntheHeavy]

\(\displaystyle Easiest \ way, \ I \ think.\)

\(\displaystyle Let \ u \ = \ ln|x|, \ \implies \ du \ = \ \frac{1}{x}dx \ and \ dv \ = \ x^{4}dx \ \implies \ v \ = \ \frac{x^{5}}{5}.\)

\(\displaystyle Hence, \ \int x^{4}ln|x|dx \ = \ \frac{x^{5}ln|x|}{5} -\int\bigg(\frac{x^{5}}{5}\bigg)\bigg(\frac{1}{x}\bigg)dx\)

\(\displaystyle = \ \frac{x^{5}}{25}[ln|x^{5}|-1]+C\)

 

[JuicyBurger]

u = ln x du = 1/x dx
v = x^5/5 dv = x^4 dx
ln x (x^5/5) - 1/5 (x^5/x) dx
1/5(x^5ln x) - 1/25x^5 + C

This is the easiest way to do this problem, in my opinion.

\(\displaystyle Let \ u = lnx \ \ \Rightarrow \ \ du = \frac{1}{x}dx\)

\(\displaystyle Let \ dv = x^4 \ \ \Rightarrow \ \ v = \frac{x^5}{5}\)

\(\displaystyle \int_{}^{}udv = uv - \int_{}^{}vdu\)

\(\displaystyle \int_{}^{}x^4ln(x)dx = \frac{x^5}{5}ln(x) - \int_{}^{}\frac{1}{x}*\frac{x^5}{5}dx = \frac{x^5}{5}ln(x) - \int_{}^{}\frac{x^4}{5}dx\)

\(\displaystyle = \frac{x^5}{5}ln(x) - \frac{x^5}{25} + C\)

The way you want to pick u and dv is by looking at what you are solving.

When you pick a u, you are going to have to find du (the derivative) and therefore want to pick something that has an easy derivative, or a simpler derivative.

When you pick a dv, you are going to have to find its integral, v, and therefore want to pick something that has an easy integral.

In our case we have \(\displaystyle \int_{}^{}x^4ln(x)dx\) which has a \(\displaystyle x^4\) and a \(\displaystyle ln(x)\).

In this case we know the derivative of \(\displaystyle ln(x)\) is \(\displaystyle \frac{1}{x}\), which is quite easy and also much simpler than it's integral, \(\displaystyle xln(x) - x\). For this reason we are going to choose \(\displaystyle ln(x)\) to be our u and \(\displaystyle x^4\) to be our dv.

Hope this helps!