Integral Calculus Question 160
Question: $ \int{32x^{3}{{(\log x)}^{2}}dx} $ is equal to:
Options:
A) $ 8x^{4}{{(\log x)}^{2}}+C $
B) $x^4 \left[ 8 (\log |x|)^2 - 4 (\log |x|) + 1 \right] + C$
C) $ x^{4}{8{{(\log x)}^{2}}-4(\log x)}+C $
D) $ x^{3}{{{(\log x)}^{2}}-2\log x}+C $
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Answer:
Correct Answer: B
Solution:
[b] Let $ I=\int{32x^{3}{{(\log x)}^{2}}dx} $ $ =32{ {{(logx)}^{2}}{{\frac{x}{4}}^{4}}-\int{2\log x\frac{1}{x}.\frac{x^{4}}{4}dx} } $ $ =\frac{32}{4}x^{4}{{(\log x)}^{2}}-16\int{x^{3}\log xdx} $ $ =8x^{4}{{(\log x)}^{2}}-4x^{4}\log x+4\int{x^{3}dx} $ $ =x^{4}{8{{(\log x)}^{2}}-4\log x+1}+C $
BETA
