Question

$\int (lnx{)}^{-1}dx-\int (lnx{)}^{-2}dx$ is equal to

• A. $x(lnx{)}^{-1}+c$
• B. $x(lnx{)}^{-2}+c$
• C. $x\left(lnx\right)+c$
• D. $x(lnx{)}^2+c$

Answer 1

Answer: (A)

$x(lnx{)}^{-1}+c$

Consider the indefinite integral

$\int (\ln x{)}^{-1}dx$

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

Applying integration by parts,

$\int \left(\ln x{)}^{-1}\right(1)dx=(\ln x{)}^{-1}\int 1dx-\int [\frac{d}{dx}(\ln x{)}^{-1}\int 1dx]dx$

$=x(\ln x{)}^{-1}-\int [\frac{-1}{(\ln x{)}^2}\frac{1}{x}x]dx+c$

$=x(\ln x{)}^{-1}-\int [\frac{-1}{(\ln x{)}^2}\frac{1}{x}x]dx+c$

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

$\therefore \int (\ln x{)}^{-1}dx=x(\ln x{)}^{-1}+\int (\ln x{)}^{-2}dx+c$

$\Rightarrow \int (\ln x{)}^{-1}dx-\int (\ln x{)}^{-2}dx=x(\ln x{)}^{-1}+c$

The answer: Option A.