Question

The value of the indefinite integral $\int{32{{x}^{3}}{{\left( \log x \right)}^{2}}dx}$ is equal to:
$\left( \text{a} \right)\text{ }8{{x}^{4}}{{\left( \log x \right)}^{2}}+C$
$\left( \text{b} \right)\text{ }{{x}^{4}}\left[ 8{{\left( \log x \right)}^{2}}-4\log x+1 \right]+C$
$\left( \text{c} \right)\text{ }{{x}^{4}}8{{\left( \log x \right)}^{2}}-4\log x+C$
$\left( \text{d} \right)\text{ }{{x}^{3}}{{\left( \log x \right)}^{2}}+2\log x+C$

Answer 1

Hint: To solve the given question, we will first assume that the value of the given integral is I. Then, we will find the value of $\int{{{x}^{3}}\log xdx}$ using the method of by – parts. According to the method of by – parts, we have $\int{u.v=v\int{udx}-\int{\dfrac{dv}{dx}\int{u{{\left( dx \right)}^{2}}.}}}$ In our case, we will take $u={{x}^{3}}$ and v = log x. After finding this, we will find the value of $\int{{{x}^{3}}{{\left( \log x \right)}^{2}}dx}$ using by – parts. In this, we will take $u={{x}^{3}}\log x$ and v = log x. After finding the value of $\int{{{x}^{3}}{{\left( \log x \right)}^{2}}dx}$ we will put its value in I and then we will simplify to get the answer.

Complete step-by-step answer:
To start with, we will assume that the value of the given indefinite integral is I. Thus, we will get the following equation
$I=\int{32{{x}^{3}}{{\left( \log x \right)}^{2}}dx}$
$\Rightarrow I=32\int{{{x}^{3}}{{\left( \log x \right)}^{2}}dx.....\left( i \right)}$
Now, the first thing we are going to do is to find the value of $\int{{{x}^{3}}\log xdx.}$ Let its value be A. Thus, we have,
$A=\int{{{x}^{3}}\log xdx.....\left( ii \right)}$
Now, we will find the value of A by using it by parts. By – parts is the method of integration when the terms inside the integration are in multiplication. Thus, we have,
$\int{u.vdx=v\int{udx-\int{\dfrac{dv}{dx}\int{u{{\left( dx \right)}^{2}}}}}}$
In our case, we will take $u={{x}^{3}}$ and v = log x. Thus, we will get,
$\int{{{x}^{3}}\log xdx=\log x\int{{{x}^{3}}dx-\int{\dfrac{d}{dx}\left( \log x \right)\int{{{x}^{3}}{{\left( dx \right)}^{2}}}}}}$
The value of $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}$ and differentiation of log x is $\dfrac{1}{x}.$ Thus, we have,
$\int{{{x}^{3}}\log xdx=\dfrac{{{x}^{4}}\log x}{4}-\int{\dfrac{1}{x}\left[ \dfrac{{{x}^{4}}}{4} \right]dx+{{C}_{1}}}}$
$\Rightarrow \int{{{x}^{3}}\log xdx=\dfrac{{{x}^{4}}\log x}{4}-\int{\dfrac{{{x}^{3}}}{4}dx+{{C}_{1}}}}$
$\Rightarrow \int{{{x}^{3}}\log xdx=\dfrac{{{x}^{4}}\log x}{4}-\dfrac{{{x}^{4}}}{4\times 4}+{{C}_{1}}}$
$\Rightarrow \int{{{x}^{3}}\log xdx=\dfrac{{{x}^{4}}\log x}{4}-\dfrac{{{x}^{4}}}{16}+{{C}_{1}}.....\left( iii \right)}$
Now, we will find the value of $\int{{{x}^{3}}{{\left( \log x \right)}^{2}}dx}$ by the help of by – parts method. Here, we will take $u={{x}^{3}}\log x$ and v = log x. Thus, we have,
$\int{{{x}^{3}}{{\left( \log x \right)}^{2}}dx=\log x\int{{{x}^{3}}\log xdx}-\int{\dfrac{d}{dx}\left( \log x \right)}\int{{{x}^{3}}\log x{{\left( dx \right)}^{2}}}}$
Now, we will put the value of $\int{{{x}^{3}}\log xdx}$ from (iii) to the above equation. Thus, we will get,
$\Rightarrow \int{{{x}^{3}}{{\left( \log x \right)}^{2}}dx}=\log x\left[ \dfrac{{{x}^{4}}\log x}{4}-\dfrac{{{x}^{4}}}{16} \right]-\int{\dfrac{1}{x}\left[ \dfrac{{{x}^{4}}\log x}{4}-\dfrac{{{x}^{4}}}{16} \right]dx+C}$
$\Rightarrow \int{{{x}^{3}}{{\left( \log x \right)}^{2}}dx}=\dfrac{{{x}^{4}}{{\left( \log x \right)}^{2}}}{4}-\dfrac{{{x}^{4}}\log x}{16}-\int{\dfrac{{{x}^{3}}\log xdx}{4}}+\int{\dfrac{{{x}^{3}}}{16}dx}+C$
Now, we will put the value of $\int{{{x}^{3}}\log xdx}$ in the above equation. Thus, we will get,
$\Rightarrow \int{{{x}^{3}}{{\left( \log x \right)}^{2}}dx}=\dfrac{{{x}^{4}}{{\left( \log x \right)}^{2}}}{4}-\dfrac{{{x}^{4}}\log x}{16}-\dfrac{1}{4}\left[ \dfrac{{{x}^{4}}\log x}{4}-\dfrac{{{x}^{4}}}{16} \right]+\dfrac{{{x}^{4}}}{4\times 16}+C$
$\Rightarrow \int{{{x}^{3}}{{\left( \log x \right)}^{2}}dx}=\dfrac{{{x}^{4}}}{4}\left[ {{\left( \log x \right)}^{2}}-\dfrac{\log x}{4} \right]-\dfrac{{{x}^{4}}\log x}{16}+\dfrac{{{x}^{4}}}{64}+\dfrac{{{x}^{4}}}{64}+C$
$\Rightarrow \int{{{x}^{3}}{{\left( \log x \right)}^{2}}dx}=\dfrac{{{x}^{4}}{{\left( \log x \right)}^{2}}}{4}-\dfrac{{{x}^{4}}\log x}{8}+\dfrac{{{x}^{4}}}{32}+C$
$\Rightarrow \int{{{x}^{3}}{{\left( \log x \right)}^{2}}dx}=\dfrac{{{x}^{4}}}{32}\left[ 8{{\left( \log x \right)}^{2}}-4\left( \log x \right)+1 \right]+C.....\left( iv \right)$
On putting the values of $\int{{{x}^{3}}{{\left( \log x \right)}^{2}}dx}$ from (iv) to (i), we will get,
I=32\left[ \dfrac{{{x}^{4}}}{32}\left\\{ 8{{\left( \log x \right)}^{2}}-4\left( \log x \right)+1 \right\\} \right]+C
$I={{x}^{4}}\left[ 8{{\left( \log x \right)}^{2}}-4\log x+1 \right]+C$
Hence, option (b) is the right answer.

Note: We can also solve the question by differentiating each option and the option whose differentiation will be equal to $32{{x}^{3}}{{\left( \log x \right)}^{2}}$ will be the answer. For instance, the differentiation of the term in option (b) is:
\dfrac{d}{dx}\left[ {{x}^{4}}\left\\{ 8{{\left( \log x \right)}^{2}}-4\left( \log x \right)+1 \right\\} \right]=8\dfrac{d}{dx}\left[ {{x}^{4}}{{\left( \log x \right)}^{2}} \right]-4\dfrac{d}{dx}\left[ {{x}^{4}}\log x \right]+\dfrac{d}{dx}{{x}^{4}}
\Rightarrow \dfrac{d}{dx}\left[ {{x}^{4}}\left\\{ 8{{\left( \log x \right)}^{2}}-4\left( \log x \right)+1 \right\\} \right]=8\left[ 4{{x}^{3}}{{\left( \log x \right)}^{2}}+\dfrac{2{{x}^{4}}}{x}\log x \right]-4\left[ 4{{x}^{3}}\log x+\dfrac{{{x}^{4}}}{x} \right]+4{{x}^{3}}
\Rightarrow \dfrac{d}{dx}\left[ {{x}^{4}}\left\\{ 8{{\left( \log x \right)}^{2}}-4\left( \log x \right)+1 \right\\} \right]=32{{x}^{3}}{{\left( \log x \right)}^{2}}+16{{x}^{3}}\log x-16{{x}^{3}}\log x-4{{x}^{3}}+4{{x}^{3}}
\Rightarrow \dfrac{d}{dx}\left[ {{x}^{4}}\left\\{ 8{{\left( \log x \right)}^{2}}-4\left( \log x \right)+1 \right\\} \right]=32{{x}^{3}}{{\left( \log x \right)}^{2}}
Hence, we get the right answer as option (b).