Q. $\int_{}^{}{32x^{3}\left( \log x \right)^{2}dx}$ is equal to

A) $8x^{4}\left( \log x \right)^{2} + c$
B) $x^{4}\left\{ 8\left( \log x \right)^{2} - 4\log{x + 1} \right\} + c$
C) $x^{4}\left\{ 8\left( \log x \right)^{2} - 4\log x \right\} + c$
D) $x^{3}\left\{ \left( \log x \right)^{2} - 2\log x \right\} + c$
Explanation

(b)
Let $I = \int_{}^{}{32x^{3}}\left( \log x \right)^{2}\ dx$
$= 32\ \left\{ \left( \log x \right)^{2}\frac{x^{4}}{4} - \int_{}^{}{2\log x.\frac{1}{x}.\frac{x^{4}}{4}}dx \right\}$
$= 8x^{4}\left( \log x \right)^{2} - 16\left\{ \log{x.}\frac{x^{4}}{4} - \int_{}^{}{\frac{1}{x}.\frac{x^{4}}{4}dx} \right\}$
$= 8x^{4}\left( \log x \right)^{2} - 4x^{4}\log{x + x^{4}} + c$
$= x^{4}\left\lbrack 8\ \left( \log x \right)^{2} - 4\log{x + 1} \right\rbrack + c$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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