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7.1 inderfinite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.1 inderfinite integral1 Mark
MCQ
$\int_{}^{} {32{x^3}{{(\log x)}^2}dx} $ is equal to
  • ✓
    ${x^4}\{ 8{(\log x)^2} - 4(\log x) + 1\} + c$
  • B
    ${x^3}\{ {(\log x)^2} + 2\log x\} + c$
  • C
    ${x^4}\{ 8{(\log x)^2} - 4\log x\} + c$
  • D
    $8{x^4}{(\log x)^2} + c$
✓

Answer

Correct option: A.
${x^4}\{ 8{(\log x)^2} - 4(\log x) + 1\} + c$
a
(a) 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}} {\rm{ }}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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