$\int \log x \cdot(\log x+2) d x=$
$\int \log x \cdot(\log x+2) d x=$
- $e^{x}(\log x)^{2}+c$
- $(\log x)^{2}+c$
- $x(\log x)^{2}+c$
- $x \log x+c$
Solution
Let $\mathrm{I}=\int \log \mathrm{x} \cdot(\log \mathrm{x}+2) \mathrm{dx}$
Put $\quad \log \mathrm{x}=t \Rightarrow \frac{1}{\mathrm{x}} \mathrm{dx}=\mathrm{dt} \Rightarrow \mathrm{dx}=\mathrm{e}^{t} \mathrm{dt}$
$\therefore \begin{aligned} \mathrm{I} &=\int \mathrm{e}^{\mathrm{t}}[\mathrm{t}(\mathrm{t}+2)] \mathrm{dt} \\ &=\int \mathrm{e}^{t}\left(\mathrm{t}^{2}+2 \mathrm{t}\right) \mathrm{dt}=\mathrm{e}^{t}\left(\mathrm{t}^{2}\right)+\mathrm{c} \\ &=\mathrm{x}[\log \mathrm{x}]^{2}+\mathrm{c} \end{aligned}$
Asked in: MHT CET 2020 (19 Oct Shift 2)
Practice more Indefinite Integration questions on Aicharya