A) \[8{{x}^{4}}{{(\log x)}^{2}}+C\]
B) \[{{x}^{4}}\{8{{(\log x)}^{2}}-4(\log x)+1\}+C\]
C) \[{{x}^{4}}\{8{{(\log x)}^{2}}-4(\log x)\}+C\]
D) \[{{x}^{3}}\{{{(\log x)}^{2}}-2\log x\}+C\]
Correct Answer: B
Solution :
Integration by parts is given as \[\int{\underset{I}{\mathop{u}}\,\underset{II}{\mathop{\text{v}}}\,}\,\,dx=u\int{\text{v}\,\text{dx-}\int{\left[ \frac{d}{dx}(u)\int{\text{v}\,\text{dx}} \right]dx}}\] Let \[I=\int{32{{x}^{3}}}\,{{(\log x)}^{2}}dx\] Integrate it by parts, using ILATE so, we choose \[{{(\log x)}^{2}}\] as \[{{I}^{st}}\] function and \[{{x}^{3}}\] as \[I{{I}^{nd}}\] function \[=32\left\{ {{(\log x)}^{2}}\frac{{{x}^{4}}}{4}-\int{2\log x\frac{1}{x}.{{\frac{x}{4}}^{4}}dx} \right\}\] \[=\frac{32}{4}{{x}^{4}}{{(\log x)}^{2}}-16\int{{{x}^{3}}\,\,\log \,x\,\,dx}\] \[=8{{x}^{4}}{{(\log \,x)}^{2}}-16\left\{ \log x.\frac{{{x}^{4}}}{4}-\int{\frac{1}{x}.\frac{{{x}^{4}}}{4}dx} \right\}\] \[=8{{x}^{4}}{{(\log \,x)}^{2}}-4{{x}^{4}}\log x+4\int{{{x}^{3}}dx}\] \[=8{{x}^{4}}{{(\log x)}^{2}}-4{{x}^{4}}\log x+{{x}^{4}}+C\] \[{{x}^{4}}\left\{ 8{{(\log x)}^{2}}-4\log x+1 \right\}+C\]
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