How do you find the integral of #x(ln x)^3 dx#?

Question

Emilia Aguilar · Accepted Answer

I found:
#x^2/2ln^3(x)-3/4x^2ln^2(x)+3/4x^2ln(x)-3/8x^2+c#

I started integrating by Substitution and the by Parts (three times):

Cameron Alexander · Answer

This is of course Integration by Parts. Let:
#u = ln^3x#
#du = (3ln^2x)/xdx#
#dv = xdx#
#v = x^2/2# #uv - intvdu# #= x^2/2ln^3x - int x^cancel(2)/2 * (3ln^2x)/cancel(x)dx# #= (x^2ln^3x)/2 - 3/2int xln^2xdx# Repeat:
#u = ln^2x#
#du = (2lnx)/xdx#
#dv = xdx#
#v = x^2/2# #= (x^2ln^3x)/2 - 3/2(int xln^2xdx)# #= (x^2ln^3x)/2 - 3/2((x^2ln^2x)/2 - int x^cancel(2)/cancel(2) * (cancel(2)lnx)/cancel(x) dx)# #= (x^2ln^3x)/2 - 3/2((x^2ln^2x)/2 - int xlnx dx)# and repeat again:
#u = lnx#
#du = 1/xdx#
#dv = xdx#
#v = x^2/2# #= (x^2ln^3x)/2 - 3/4x^2ln^2x + 3/2(int xlnx dx)# #= (x^2ln^3x)/2 - 3/4x^2ln^2x + 3/2((x^2lnx)/2 - int x^cancel(2)/2*1/cancel(x)dx)# #= (x^2ln^3x)/2 - 3/4x^2ln^2x + 3/2((x^2lnx)/2 - int x/2dx)# #= 1/2x^2ln^3x - 3/4x^2ln^2x + 3/4x^2lnx - 3/4int xdx# #= 1/2x^2ln^3x - 3/4x^2ln^2x + 3/4x^2lnx - 3/8x^2# #= color(blue)(1/8x^2(4ln^3x - 6ln^2x + 6lnx - 3) + C)#

Emilia Aguilar · Answer

undefined To find the integral of $x(\ln x)^3$ with respect to $x$, you can use integration by parts. Let $u = \ln^3 x$ and $dv = x dx$. Then, $du = 3(\ln^2 x)(\frac{1}{x}) dx$ and $v = \frac{1}{2}x^2$.

Apply the integration by parts formula:

$$
\int u \, dv = uv - \int v \, du
$$

Substitute the values:

$$
\int x(\ln x)^3 \, dx = \frac{1}{2}x^2(\ln x)^3 - \int \frac{1}{2}x^2 \cdot 3(\ln^2 x)(\frac{1}{x}) \, dx
$$

Simplify:

$$
\int x(\ln x)^3 \, dx = \frac{1}{2}x^2(\ln x)^3 - \frac{3}{2}\int x(\ln^2 x) \, dx
$$

Now, apply integration by parts again to the remaining integral $\int x(\ln^2 x) \, dx$.

Let $u = \ln^2 x$ and $dv = x dx$. Then, $du = 2\ln x \frac{1}{x} dx$ and $v = \frac{1}{2}x^2$.

Apply integration by parts:

$$
\int x(\ln^2 x) \, dx = \frac{1}{2}x^2(\ln^2 x) - \int \frac{1}{2}x^2 \cdot 2\ln x \frac{1}{x} \, dx
$$

Simplify:

$$
\int x(\ln^2 x) \, dx = \frac{1}{2}x^2(\ln^2 x) - \int x \ln x \, dx
$$

The integral $\int x \ln x \, dx$ can be solved using integration by parts again.

Let $u = \ln x$ and $dv = x dx$. Then, $du = \frac{1}{x} dx$ and $v = \frac{1}{2}x^2$.

Apply integration by parts:

$$
\int x \ln x \, dx = \frac{1}{2}x^2 \ln x - \int \frac{1}{2}x^2 \cdot \frac{1}{x} \, dx
$$

Simplify:

$$
\int x \ln x \, dx = \frac{1}{2}x^2 \ln x - \frac{1}{4}x^2
$$

Now, substitute this result back into the integral expression:

$$
\int x(\ln^2 x) \, dx = \frac{1}{2}x^2(\ln^2 x) - \left(\frac{1}{2}x^2 \ln x - \frac{1}{4}x^2\right)
$$

Simplify further:

$$
\int x(\ln^2 x) \, dx = \frac{1}{2}x^2(\ln^2 x) - \frac{1}{2}x^2 \ln x + \frac{1}{4}x^2
$$

Finally, substitute this expression back into the original integral:

$$
\int x(\ln x)^3 \, dx = \frac{1}{2}x^2(\ln x)^3 - \frac{3}{2}\left(\frac{1}{2}x^2(\ln^2 x) - \frac{1}{2}x^2 \ln x + \frac{1}{4}x^2\right)
$$

Simplify to get the final result.