Question

Find $\int \frac{(x-1)(x-logx{)}^3}{x}dx$.

Answer 1

To find :

$I=\int \frac{{(x-1)(x-\log x)}^3}{x}dx⟶e{q}^{\prime }n-1$

Let $(x-\log x)=t$ $(1-\frac{1}{x})dx=dt$ $\left(\frac{x-1}{x}\right)dx=dt$

Now substituting values in eq’n-1

$\Rightarrow I=\int \left(\frac{x-1}{x}\right)dx{(x-\log x)}^3$

$\Rightarrow I=\int dt.t^3$

$\Rightarrow I=\int t^3.dt=\frac{t^4}{4}+C$

$\Rightarrow I=\frac{{(x-\log x)}^4}{4}+C$ (As$(x-\log x)=t$)