Integral Calculus Question 116
Question: $ \int_{{}}^{{}}{32x^{3}{{(\log x)}^{2}}dx} $ is equal to
[MP PET 2004]
Options:
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) $ 8x^{4}{{(\log x)}^{2}}+c $
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Answer:
Correct Answer: A
Solution:
Let $ I=\int_{{}}^{{}}{32x^{3}{{(\log x)}^{2}}}dx=32\int_{{}}^{{}}{x^{3}{{(\log x)}^{2}}dx} $
$ =32,[ {{(\log x)}^{2}}\int_{{}}^{{}}{x^{3}dx-\int_{{}}^{{}}{( \frac{d}{dx}{{(\log x)}^{2}}\int_{{}}^{{}}{x^{3}dx} ),dx}} ] $
$ =32,[ {{(\log x)}^{2}}.\frac{x^{4}}{4}-\int_{{}}^{{}}{2\log x.\frac{1}{x}.\frac{x^{4}}{4}dx} ] $
$ =32[ {{(\log x)}^{2}}\frac{x^{4}}{4}-\frac{1}{2}\int_{{}}^{{}}{x^{3}\log x,dx} ] $
$ =32[ \frac{{{(\log x)}^{2}}x^{4}}{4}-\frac{1}{2}( \frac{\log x.x^{4}}{4}-\int_{{}}^{{}}{\frac{1}{x}.\frac{x^{4}}{4}}dx ) ] $
$ =32[ \frac{{{(\log x)}^{2}}x^{4}}{4}-\frac{1}{2}( \frac{x^{4}\log x}{4}-\frac{1}{4}.\frac{x^{4}}{4} ) ]+c $
$ =8,[ {{(\log x)}^{2}}x^{4}-\frac{1}{2}( x^{4}\log x-\frac{x^{4}}{4} ) ]+c $
$ =8x^{4}[ {{(\log x)}^{2}}-\frac{\log x}{2}+\frac{1}{8} ]+c $
$ =x^{4}[8{{(\log x)}^{2}}-4\log x+1]+c $ .
BETA
