Question

Question #5266a

Answer 1

`-1/(3(ln(x))^3)+C`

Explanation:

`I=int1/(x(ln(x))^4)dx`

We can use substitution `u=ln(x)`. Differentiating this shows that `(du)/dx=1/x`, or `du=1/xdx`. This is already present in our integral, but we can make it clearer:

`I=int1/(ln(x))^4(1/xdx)=intunderbrace((ln(x))^-4)_(u^-4)overbrace((1/xdx))^(du)=intu^-4du`

We can now use the integration rule `intu^ndu=u^(n+1)/(n+1)+C`:

`I=u^(-4+1)/(-4+1)+C=u^(-3)/(-3)+C=-1/(3u^3)+C`

Since `u=ln(x)`:

`I=-1/(3(ln(x))^3)+C`