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

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

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

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

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

Want more practice?
Practice Integrals MCQs