A) \[{{x}^{4}}\{8{{(\log x)}^{2}}-4(\log x)+1\}+c\]
B) \[{{x}^{3}}\{{{(\log x)}^{2}}+2\log x\}+c\]
C) \[{{x}^{4}}\{8{{(\log x)}^{2}}-4\log x\}+c\]
D) \[8{{x}^{4}}{{(\log x)}^{2}}+c\]
Correct Answer: A
Solution :
Let \[I=\int_{{}}^{{}}{32{{x}^{3}}{{(\log x)}^{2}}}dx=32\int_{{}}^{{}}{{{x}^{3}}{{(\log x)}^{2}}dx}\] \[=32\,\left[ {{(\log x)}^{2}}\int_{{}}^{{}}{{{x}^{3}}dx-\int_{{}}^{{}}{\left( \frac{d}{dx}{{(\log x)}^{2}}\int_{{}}^{{}}{{{x}^{3}}dx} \right)\,dx}} \right]\] \[=32\,\left[ {{(\log x)}^{2}}.\frac{{{x}^{4}}}{4}-\int_{{}}^{{}}{2\log x.\frac{1}{x}.\frac{{{x}^{4}}}{4}dx} \right]\] \[=32\left[ {{(\log x)}^{2}}\frac{{{x}^{4}}}{4}-\frac{1}{2}\int_{{}}^{{}}{{{x}^{3}}\log x\,dx} \right]\] \[=32\left[ \frac{{{(\log x)}^{2}}{{x}^{4}}}{4}-\frac{1}{2}\left( \frac{\log x.{{x}^{4}}}{4}-\int_{{}}^{{}}{\frac{1}{x}.\frac{{{x}^{4}}}{4}}\text{ }dx \right) \right]\] \[=32\left[ \frac{{{(\log x)}^{2}}{{x}^{4}}}{4}-\frac{1}{2}\left( \frac{{{x}^{4}}\log x}{4}-\frac{1}{4}.\frac{{{x}^{4}}}{4} \right) \right]+c\] \[=8\,\left[ {{(\log x)}^{2}}{{x}^{4}}-\frac{1}{2}\left( {{x}^{4}}\log x-\frac{{{x}^{4}}}{4} \right) \right]+c\] \[=8{{x}^{4}}\left[ {{(\log x)}^{2}}-\frac{\log x}{2}+\frac{1}{8} \right]+c\] \[={{x}^{4}}[8{{(\log x)}^{2}}-4\log x+1]+c\].
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