Question

$\int (\frac{1}{\log x}-\frac{1}{(\log x{)}^2})dx$_____ $+c$

• A. $x\log .x$
• B. $\frac{x}{2}(\log x{)}^2$
• C. $\frac{x}{\log x}$
• D. $\frac{x}{(\log x{)}^2}$

Answer 1

The correct option is C $\frac{x}{\log x}$

Let $z=\log x⟹dz=\frac{dx}{x}$

$⟹dx=e^{z}dz$

Integration becomes-

$\int (\frac{1}{z}-\frac{1}{z^2})e^{z}dz$

Now, we know that

$\int \left(f\right(x)+f^{\prime }(x\left)\right)e^{x}dx=f\left(x\right)e^x+c$

Here, $f\left(z\right)=\frac{1}{z}⟹f^{\prime }\left(z\right)=\frac{-1}{z^2}$

Hence, given integration also becomes

$\int \left(f\right(z)+f^{\prime }(z\left)\right)e^{z}dz$

$=f\left(z\right)e^z+c$

$=\frac{e^z}{z}+c$

$=\frac{x}{\log x}+c$