How do you integrate #(5 ln(x))/x^(7)#?

Question

Christopher Agresta · Accepted Answer

The answer is #=-5/6lnx/x^6-5/(36x^6)+C#

Apply the Integration by parts

#intuv'=uv-u'v#

The integral is

#I=int(5lnxdx)/x^7=5int(lnx)/x^7#

#u=lnx#, #=>#, #u'=1/x#

#v'=x^-7#, #=>#, #v=-1/(6x^6)#

Therefore,

#I=-5/6lnx/x^6+5/6int(dx)/x^7#

#=-5/6lnx/x^6-5/(36x^6)+C#

Christopher Allers · Answer

undefined To integrate $ \frac{5 \ln(x)}{x^7} $, you can use integration by parts. Let $ u = \ln(x) $ and $ dv = \frac{5}{x^7}dx $. Then, $ du = \frac{1}{x}dx $ and $ v = -\frac{1}{6x^6} $.

Applying the integration by parts formula:

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

Substituting the values:

$$ \int \frac{5 \ln(x)}{x^7} \, dx = -\frac{5 \ln(x)}{6x^6} - \int \left(-\frac{1}{6x^6}\right) \cdot \frac{1}{x} \, dx $$

Simplify the integral:

$$ \int \frac{5 \ln(x)}{x^7} \, dx = -\frac{5 \ln(x)}{6x^6} + \frac{1}{6} \int \frac{1}{x^7} \, dx $$

Now, integrate $ \frac{1}{x^7} $:

$$ \int \frac{1}{x^7} \, dx = \frac{-1}{6x^6} + C $$

So, putting it all together:

$$ \int \frac{5 \ln(x)}{x^7} \, dx = -\frac{5 \ln(x)}{6x^6} - \frac{1}{6} \left(\frac{-1}{6x^6}\right) + C $$

$$ = -\frac{5 \ln(x)}{6x^6} + \frac{1}{36x^6} + C $$

Where $ C $ is the constant of integration.