How do you integrate #int lnx/x# by integration by parts method?

Question

Luke Alvarez · Accepted Answer

#=>I=(1/2)(lnx)^2+K#

integration by parts formula:

#intu((dv)/dx)dx=uv-intv((du)/dx)dx#

for #color(red)(I=(int(lnx/x)dx))#

Let #u= lnx=>(du)/dx=1/x#

Let #(dv)/dx=1/x=>v=lnx#

#:. I=(lnx)(lnx)-color(red)(int(lnx/x)dx)#

#=>I=(lnx)^2-color(red)(I)+C#

#=>2I=(lnx)^2+C#

#=>I=(1/2)(lnx)^2+K#

Samantha Adkison · Answer

# int lnx /x dx = 1/2(lnx)^2 + c #

The formula for IBP is # int u (dv)/dx dx= uv- int v(du)/dx dx #

Se hopefully we can identify one function which simplifies when differentiated, and the other which simplifies when integrated (or is at least easier to integrate than the original function).

so hopefully it is obvious that we choose: #u=lnx# and #(du)/dx=1/x#

so #u=lnx=>(du)/dx=1/x# and #(dv)/dx=1/x=>v=lnx#

# :. int lnx 1/x dx = lnxlnx-int lnx1/xdx + c_1# # :. int lnx /x dx = (lnx)^2-int lnx/xdx + c_1# # :. 2int lnx /x dx = (lnx)^2 + c_1# # :. int lnx /x dx = 1/2(lnx)^2 + 1/2c_1# # :. int lnx /x dx = 1/2(lnx)^2 + c #

Luke Alvarez · Answer

undefined To integrate $ \int \frac{\ln(x)}{x} \, dx $ using integration by parts, use the formula $\int u \, dv = uv - \int v \, du$. Let:

- $ u = \ln(x) $, which means $ du = \frac{1}{x} \, dx $.
- $ dv = \frac{1}{x} \, dx $, which implies $ v = \ln(x) $.

Now, apply the integration by parts formula:

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

Here, we notice that the integral we're trying to solve appears on both sides of the equation, which was a mistake in choosing $u$ and $dv$. The correct setup for integration by parts in this case should actually not involve integration by parts directly, as my initial setup leads to a circular problem. Let's correct this approach:

Given the integral:
$$ \int \frac{\ln(x)}{x} \, dx $$

We recognize this integral can be directly integrated without using the method of integration by parts due to its specific form. It's a common integral that results in a known function:

The correct integration is:
$$ \int \frac{\ln(x)}{x} \, dx = \frac{1}{2} (\ln(x))^2 + C $$

This result is obtained because the integral is a standard form where integrating $\ln(x)/x$ directly results in $\frac{1}{2} (\ln(x))^2 + C$, based on the recognition that the derivative of $\ln(x)$ is $1/x$, making the integral of $\frac{\ln(x)}{x}$ take this form after applying the power rule for integration indirectly. My initial direction was incorrect; for this specific integral, direct recognition or substitution is the right approach, not integration by parts.