Question

If $\int \left[\log \left(\log \left(x\right)\right)+\frac{1}{{\left(\log \left(x\right)\right)}^2}\right]dx=x\left[f\left(x\right)-g\left(x\right)\right]+c$then

• A. $f\left(x\right)=\log \left(\log \left(x\right)\right);g\left(x\right)=\frac{1}{\log \left(x\right)}$
• B. $f\left(x\right)=\log \left(x\right);g\left(x\right)=\frac{1}{\log \left(x\right)}$
• C. $f\left(x\right)=\frac{1}{x\log \left(x\right)};g\left(x\right)=\log \left(\log \left(x\right)\right)$
• D. $f\left(x\right)=\frac{1}{x\log \left(x\right)};g\left(x\right)=\frac{1}{\log \left(x\right)}$

Answer 1

The correct option is A

$f\left(x\right)=\log \left(\log \left(x\right)\right);g\left(x\right)=\frac{1}{\log \left(x\right)}$

Explanation for correct option

Given, $\int \left[\log \left(\log \left(x\right)\right)+\frac{1}{{\left(\log \left(x\right)\right)}^2}\right]dx=x\left[f\left(x\right)-g\left(x\right)\right]+c$

Let $\log \left(x\right)=z$

$$\begin{gathered} \Rightarrow xdz=dx \\ \Rightarrow e^{z}dz=dx \\ \therefore \int \left[\log \left(\log \left(x\right)\right)+\frac{1}{{\left(\log \left(x\right)\right)}^2}\right]dx \\ =\int \left[\log \left(z\right)+\frac{1}{z^2}\right]e^{z}dz \\ =\int e^z\log \left(z\right)dz+\int \frac{e^z}{z^2}dz \end{gathered}$$

Using integration by-parts we have

$$\begin{gathered} \Rightarrow \int e^z\log \left(z\right)dz+\int \frac{e^z}{z^2}dz \\ =\log \left(z\right)\int e^{z}dz-\int \left[\frac{d\left(\log \left(z\right)\right)}{dz}\int e^{z}dz\right]dz+\int \frac{e^z}{z^2}dz \\ =e^z\log \left(z\right)-\int \left[\frac{e^z}{z}\right]dz+\int \frac{e^z}{z^2}dz \\ =e^z\log \left(z\right)-\left[\frac{1}{z}\int \left[e^z\right]dz-\int \left[\frac{d}{dz}\left(\frac{1}{z}\right)\int e^{z}dz\right]dz\right]+\int \frac{e^z}{z^2}dz \\ =e^z\log \left(z\right)-\left[\frac{e^z}{z^{}}+\int \frac{e^z}{z^2}dz\right]+\int \frac{e^z}{z^2}dz \\ =e^z\log \left(z\right)-\frac{e^z}{z}-\int \frac{e^z}{z^2}dz+\int \frac{e^z}{z^2}dz \\ =e^z\log \left(z\right)-\frac{e^z}{z}+c \\ =x\log \left(\log \left(x\right)\right)-\frac{x}{\left(\log \left(x\right)\right)}+c\left[\because \log \left(x\right)=z,e^z=x\right] \\ =x\left[\log \left(\log \left(x\right)\right)-\frac{1}{\left(\log \left(x\right)\right)}\right]+c \end{gathered}$$

Hence, option A is correct