_ ^ 32 x^3 ( x) ^2 dx is equal to - Vidyadip

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$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

The mean and variance of a binomial distribution are $4$ and $3$ respectively, then the probability of getting exactly six successes in this distribution is

Area bounded by curves $x =\sqrt {y -1}$ and $y = x + 1$ is-

Let $f, g: R \to R$ be two functions defined by $f(x)\, = \,\left\{ {\begin{array}{*{20}{c}}{x\,\sin \,\left( {\frac{1}{x}} \right),\,x\, \ne \,0\,\,\,\,\,\,\,\,\,\,}\\ {0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,x\, = 0\,\,\,\,\,\,\,\,\,}\end{array}} \right.,$ and $g(x) =x\,f(x)$
Statement $I: f$ is a continuous function at $x = 0.$
Statement $II:$ $g$ is a differentiable function at $x = 0.$

If the sum of the coefficients of all the positive even powers of $x$ in the binomial expansion of $\left(2 x^{3}+\frac{3}{x}\right)^{10}$ is $5^{10}-\beta \cdot 3^{9}$, then $\beta$ is equal to

When $\frac{{z + i}}{{z + 2}}$ is purely imaginary, the locus described by the point $z$ in the Argand diagram is a

The parabola $y^2=4 x+1$ divides the disc $x^2+y^2 \leq 1$ into two regions with areas $A_1$ and $A_2$. Then, $\left|A_1-A_2\right|$ equals

If $x = a\sin 2\theta (1 + \cos 2\theta ),y = b\cos 2\theta (1 - \cos 2\theta )$, then ${{dy} \over {dx}} = $

If the circles $x^{2}+y^{2}+6 x+8 y+16=0$ and $x^{2}+y^{2}+2(3-\sqrt{3}) x+x+2(4-\sqrt{6}) y$ $= k +6 \sqrt{3}+8 \sqrt{6}, k >0$, touch internally at the point $P(\alpha, \beta)$, then $(\alpha+\sqrt{3})^{2}+(\beta+\sqrt{6})^{2}$ is equal to $\dots\dots$

The equation of a line passing through the point $(-3, 2, -4)$ and equally inclined to the axes, are

Six objects $O_1$ to $O_6$ are arranged one on top of the other. In how many ways can these be arranged such that $O_1$ and $O_2$ are the $2$ bottom most objects ?