If \[\int{\dfrac{7{{x}^{8}}+8{{x}^{7}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}dx=f\left( x \right)+c}\] then \[f\left( x \right)\] is equal to
(A) \[\dfrac{{{x}^{8}}}{\left( 1+x+{{x}^{8}} \right)}\]
(B) \[28\log \left( 1+x+{{x}^{8}} \right)\]
(C) \[\dfrac{1}{\left( 1+x+{{x}^{8}} \right)}\]
(D) \[\dfrac{-1}{\left( 1+x+{{x}^{8}} \right)}\]
Answer
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Hint: First of all, in the denominator, divide and multiply by the term \[{{x}^{16}}\] . Now, simplify it as \[\int{\dfrac{\dfrac{7{{x}^{8}}}{{{x}^{16}}}+\dfrac{8{{x}^{7}}}{{{x}^{16}}}}{{{\left( \dfrac{1}{{{x}^{8}}}+\dfrac{1}{{{x}^{7}}}+1 \right)}^{2}}}dx}\] . Assume that \[t=\left( \dfrac{1}{{{x}^{8}}}+\dfrac{1}{{{x}^{7}}}+1 \right)\] . Now, differentiate \[t\] with respect to \[dx\] and get the relation between \[dx\] and \[dt\] . Transform \[\int{\dfrac{\dfrac{7{{x}^{8}}}{{{x}^{16}}}+\dfrac{8{{x}^{7}}}{{{x}^{16}}}}{{{\left( \dfrac{1}{{{x}^{8}}}+\dfrac{1}{{{x}^{7}}}+1 \right)}^{2}}}dx}\] in terms of \[t\] . Use the formula \[\int{{{t}^{n}}dt=\dfrac{{{t}^{\left( n+1 \right)}}}{\left( n+1 \right)}}+c\] and solve it further. At last, replace \[t\] by \[\left( \dfrac{1}{{{x}^{8}}}+\dfrac{1}{{{x}^{7}}}+1 \right)\] and then compare it with \[f\left( x \right)+c\] to obtain \[f\left( x \right)\] .
Complete step by step answer:
According to the question, we are given that
\[\int{\dfrac{7{{x}^{8}}+8{{x}^{7}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}dx=f\left( x \right)+c}\] …………………………………….(1)
Here, we are asked to find the function \[f\left( x \right)\] .
First of all, we have to integrate the expression, \[\int{\dfrac{7{{x}^{8}}+8{{x}^{7}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}dx}\] ……………………………………..(2)
On dividing and multiplying by \[{{x}^{16}}\] in the denominator of equation (2), we get
\[\Rightarrow \int{\dfrac{7{{x}^{8}}+8{{x}^{7}}}{\dfrac{{{x}^{16}}}{{{x}^{16}}}{{\left( 1+x+{{x}^{8}} \right)}^{2}}}dx}\]
\[\Rightarrow \int{\dfrac{7{{x}^{8}}+8{{x}^{7}}}{{{x}^{16}}{{\left( \dfrac{1+x+{{x}^{8}}}{{{x}^{8}}} \right)}^{2}}}dx}\] ………………………………………(3)
Now, after further simplifying equation (3), we get
\[=\int{\dfrac{\dfrac{7{{x}^{8}}}{{{x}^{16}}}+\dfrac{8{{x}^{7}}}{{{x}^{16}}}}{{{\left( \dfrac{1}{{{x}^{8}}}+\dfrac{1}{{{x}^{7}}}+1 \right)}^{2}}}dx}\]
\[\Rightarrow \int{\dfrac{\dfrac{7}{{{x}^{8}}}+\dfrac{8}{{{x}^{9}}}}{{{\left( \dfrac{1}{{{x}^{8}}}+\dfrac{1}{{{x}^{7}}}+1 \right)}^{2}}}dx}\] ………………………………………(4)
Let us assume that \[t=\left( \dfrac{1}{{{x}^{8}}}+\dfrac{1}{{{x}^{7}}}+1 \right)\] …………………………………………(5)
On differentiating the LHS and RHS of equation (5) with respect to \[dx\] , we get
\[\begin{align}
& \Rightarrow \dfrac{dt}{dx}=-\dfrac{7}{{{x}^{8}}}-\dfrac{8}{{{x}^{9}}} \\
& \Rightarrow dt=-\left( \dfrac{7}{{{x}^{8}}}+\dfrac{8}{{{x}^{9}}} \right)dx \\
\end{align}\]
\[\Rightarrow -dt=\left( \dfrac{7}{{{x}^{8}}}+\dfrac{8}{{{x}^{9}}} \right)dx\] ……………………………………………..(6)
Now, from equation (4), equation (5), and equation (6), we get
\[\Rightarrow \int{\dfrac{-dt}{{{\left( t \right)}^{2}}}}\]
\[\Rightarrow -\int{{{t}^{-2}}dt}\] ………………………………………………(7)
We know the formula, \[\int{{{t}^{n}}dt=\dfrac{{{t}^{\left( n+1 \right)}}}{\left( n+1 \right)}}+c\] ………………………………………..(8)
Using the formula shown in equation (8) and on simplifying equation (7), we get
\[\begin{align}
& \Rightarrow -\left\{ \dfrac{{{t}^{\left( -2+1 \right)}}}{\left( -2+1 \right)} \right\}+c \\
& \Rightarrow -\left( \dfrac{{{t}^{-1}}}{-1} \right)+c \\
\end{align}\]
\[\Rightarrow \dfrac{1}{t}\] ……………………………………(9)
Using equation (5) and on replacing \[t\] by \[\left( \dfrac{1}{{{x}^{8}}}+\dfrac{1}{{{x}^{7}}}+1 \right)\] in equation (9), we get
\[\Rightarrow\dfrac{1}{\left( \dfrac{1}{{{x}^{8}}}+\dfrac{1}{{{x}^{7}}}+1 \right)}+c\] ………………………………………………..(10)
Now, on multiplying by \[{{x}^{8}}\] in numerator and denominator in equation (10), we get
\[\begin{align}
& \Rightarrow \dfrac{{{x}^{8}}\times 1}{{{x}^{8}}\times \left( \dfrac{1}{{{x}^{8}}}+\dfrac{1}{{{x}^{7}}}+1 \right)}+c \\
& \Rightarrow \dfrac{{{x}^{8}}}{\left( \dfrac{{{x}^{8}}}{{{x}^{8}}}+\dfrac{{{x}^{8}}}{{{x}^{7}}}+{{x}^{8}} \right)}+c \\
\end{align}\]
\[\Rightarrow \dfrac{{{x}^{8}}}{\left( 1+x+{{x}^{8}} \right)}+c\] …………………………………(11)
From equation (1), we have, \[\int{\dfrac{7{{x}^{8}}+8{{x}^{7}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}dx=f\left( x \right)+c}\] .
Now, on comparing equation (1) and equation (11), we get
\[\int{\dfrac{7{{x}^{8}}+8{{x}^{7}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}dx=f\left( x \right)+c}=\dfrac{{{x}^{8}}}{\left( 1+x+{{x}^{8}} \right)}+c\]
So, \[f\left( x \right)=\dfrac{{{x}^{8}}}{\left( 1+x+{{x}^{8}} \right)}\] .
So, the correct answer is “Option A”.
Note: We know that differentiation is just opposite of integration. It means that a function obtained by integrating a function is the same function which on differentiation given the function for that is to be integrated. We can also solve this question by differentiating the given options below. Let us differentiate the first option.
\[\dfrac{d}{dx}\left( \dfrac{{{x}^{8}}}{\left( 1+x+{{x}^{8}} \right)} \right)=\dfrac{d\left\{ {{x}^{8}}{{\left( 1+x+{{x}^{8}} \right)}^{-1}} \right\}}{dx}\] ………………………………(1)
We know the formula, \[\dfrac{d\left( uv \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\] ………………………………………(2)
Now, from equation (1) and equation (2), we get
\[={{x}^{8}}\dfrac{d\left\{ {{\left( 1+x+{{x}^{8}} \right)}^{-1}} \right\}}{dx}+{{\left( 1+x+{{x}^{8}} \right)}^{-1}}\dfrac{d\left( {{x}^{8}} \right)}{dx}\]
\[={{x}^{8}}\dfrac{d\left\{ {{\left( 1+x+{{x}^{8}} \right)}^{-1}} \right\}}{dx}+{{\left( 1+x+{{x}^{8}} \right)}^{-1}}\times \left( 8{{x}^{7}} \right)\] ……………………………………..(3)
Using chain rule and on simplifying equation (3), we get
\[\begin{align}
& \Rightarrow {{x}^{8}}\dfrac{d\left\{ {{\left( 1+x+{{x}^{8}} \right)}^{-1}} \right\}}{d\left( 1+x+{{x}^{8}} \right)}\times \dfrac{d\left( 1+x+{{x}^{8}} \right)}{dx}+\dfrac{1}{\left( 1+x+{{x}^{8}} \right)}\left( 8{{x}^{7}} \right) \\
& \Rightarrow {{x}^{8}}\left( -1 \right){{\left( 1+x+{{x}^{8}} \right)}^{-1-1}}\times \left( 8{{x}^{7}}+1 \right)+\dfrac{1}{\left( 1+x+{{x}^{8}} \right)}\left( 8{{x}^{7}} \right) \\
& \Rightarrow -\dfrac{{{x}^{8}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}\left( 8{{x}^{7}}+1 \right)+\dfrac{1}{\left( 1+x+{{x}^{8}} \right)}\left( 8{{x}^{7}} \right) \\
& \Rightarrow -\dfrac{{{x}^{8}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}-\dfrac{{{x}^{8}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}\left( 8{{x}^{7}} \right)+\dfrac{1}{\left( 1+x+{{x}^{8}} \right)}\left( 8{{x}^{7}} \right) \\
& \Rightarrow -\dfrac{{{x}^{8}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}+\dfrac{\left( 8{{x}^{7}} \right)}{\left( 1+x+{{x}^{8}} \right)}\left\{ -\dfrac{{{x}^{8}}}{\left( 1+x+{{x}^{8}} \right)}+1 \right\} \\
& \Rightarrow -\dfrac{{{x}^{8}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}+\dfrac{\left( 8{{x}^{7}} \right)}{\left( 1+x+{{x}^{8}} \right)}\left\{ \dfrac{1+x}{\left( 1+x+{{x}^{8}} \right)} \right\} \\
& \Rightarrow -\dfrac{{{x}^{8}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}+\dfrac{\left( 8{{x}^{7}} \right)}{\left( 1+x+{{x}^{8}} \right)}+\dfrac{8{{x}^{8}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}} \\
& \Rightarrow -\dfrac{7{{x}^{8}}+8{{x}^{7}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}} \\
\end{align}\]
Therefore, option (A) is the correct option.
Complete step by step answer:
According to the question, we are given that
\[\int{\dfrac{7{{x}^{8}}+8{{x}^{7}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}dx=f\left( x \right)+c}\] …………………………………….(1)
Here, we are asked to find the function \[f\left( x \right)\] .
First of all, we have to integrate the expression, \[\int{\dfrac{7{{x}^{8}}+8{{x}^{7}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}dx}\] ……………………………………..(2)
On dividing and multiplying by \[{{x}^{16}}\] in the denominator of equation (2), we get
\[\Rightarrow \int{\dfrac{7{{x}^{8}}+8{{x}^{7}}}{\dfrac{{{x}^{16}}}{{{x}^{16}}}{{\left( 1+x+{{x}^{8}} \right)}^{2}}}dx}\]
\[\Rightarrow \int{\dfrac{7{{x}^{8}}+8{{x}^{7}}}{{{x}^{16}}{{\left( \dfrac{1+x+{{x}^{8}}}{{{x}^{8}}} \right)}^{2}}}dx}\] ………………………………………(3)
Now, after further simplifying equation (3), we get
\[=\int{\dfrac{\dfrac{7{{x}^{8}}}{{{x}^{16}}}+\dfrac{8{{x}^{7}}}{{{x}^{16}}}}{{{\left( \dfrac{1}{{{x}^{8}}}+\dfrac{1}{{{x}^{7}}}+1 \right)}^{2}}}dx}\]
\[\Rightarrow \int{\dfrac{\dfrac{7}{{{x}^{8}}}+\dfrac{8}{{{x}^{9}}}}{{{\left( \dfrac{1}{{{x}^{8}}}+\dfrac{1}{{{x}^{7}}}+1 \right)}^{2}}}dx}\] ………………………………………(4)
Let us assume that \[t=\left( \dfrac{1}{{{x}^{8}}}+\dfrac{1}{{{x}^{7}}}+1 \right)\] …………………………………………(5)
On differentiating the LHS and RHS of equation (5) with respect to \[dx\] , we get
\[\begin{align}
& \Rightarrow \dfrac{dt}{dx}=-\dfrac{7}{{{x}^{8}}}-\dfrac{8}{{{x}^{9}}} \\
& \Rightarrow dt=-\left( \dfrac{7}{{{x}^{8}}}+\dfrac{8}{{{x}^{9}}} \right)dx \\
\end{align}\]
\[\Rightarrow -dt=\left( \dfrac{7}{{{x}^{8}}}+\dfrac{8}{{{x}^{9}}} \right)dx\] ……………………………………………..(6)
Now, from equation (4), equation (5), and equation (6), we get
\[\Rightarrow \int{\dfrac{-dt}{{{\left( t \right)}^{2}}}}\]
\[\Rightarrow -\int{{{t}^{-2}}dt}\] ………………………………………………(7)
We know the formula, \[\int{{{t}^{n}}dt=\dfrac{{{t}^{\left( n+1 \right)}}}{\left( n+1 \right)}}+c\] ………………………………………..(8)
Using the formula shown in equation (8) and on simplifying equation (7), we get
\[\begin{align}
& \Rightarrow -\left\{ \dfrac{{{t}^{\left( -2+1 \right)}}}{\left( -2+1 \right)} \right\}+c \\
& \Rightarrow -\left( \dfrac{{{t}^{-1}}}{-1} \right)+c \\
\end{align}\]
\[\Rightarrow \dfrac{1}{t}\] ……………………………………(9)
Using equation (5) and on replacing \[t\] by \[\left( \dfrac{1}{{{x}^{8}}}+\dfrac{1}{{{x}^{7}}}+1 \right)\] in equation (9), we get
\[\Rightarrow\dfrac{1}{\left( \dfrac{1}{{{x}^{8}}}+\dfrac{1}{{{x}^{7}}}+1 \right)}+c\] ………………………………………………..(10)
Now, on multiplying by \[{{x}^{8}}\] in numerator and denominator in equation (10), we get
\[\begin{align}
& \Rightarrow \dfrac{{{x}^{8}}\times 1}{{{x}^{8}}\times \left( \dfrac{1}{{{x}^{8}}}+\dfrac{1}{{{x}^{7}}}+1 \right)}+c \\
& \Rightarrow \dfrac{{{x}^{8}}}{\left( \dfrac{{{x}^{8}}}{{{x}^{8}}}+\dfrac{{{x}^{8}}}{{{x}^{7}}}+{{x}^{8}} \right)}+c \\
\end{align}\]
\[\Rightarrow \dfrac{{{x}^{8}}}{\left( 1+x+{{x}^{8}} \right)}+c\] …………………………………(11)
From equation (1), we have, \[\int{\dfrac{7{{x}^{8}}+8{{x}^{7}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}dx=f\left( x \right)+c}\] .
Now, on comparing equation (1) and equation (11), we get
\[\int{\dfrac{7{{x}^{8}}+8{{x}^{7}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}dx=f\left( x \right)+c}=\dfrac{{{x}^{8}}}{\left( 1+x+{{x}^{8}} \right)}+c\]
So, \[f\left( x \right)=\dfrac{{{x}^{8}}}{\left( 1+x+{{x}^{8}} \right)}\] .
So, the correct answer is “Option A”.
Note: We know that differentiation is just opposite of integration. It means that a function obtained by integrating a function is the same function which on differentiation given the function for that is to be integrated. We can also solve this question by differentiating the given options below. Let us differentiate the first option.
\[\dfrac{d}{dx}\left( \dfrac{{{x}^{8}}}{\left( 1+x+{{x}^{8}} \right)} \right)=\dfrac{d\left\{ {{x}^{8}}{{\left( 1+x+{{x}^{8}} \right)}^{-1}} \right\}}{dx}\] ………………………………(1)
We know the formula, \[\dfrac{d\left( uv \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\] ………………………………………(2)
Now, from equation (1) and equation (2), we get
\[={{x}^{8}}\dfrac{d\left\{ {{\left( 1+x+{{x}^{8}} \right)}^{-1}} \right\}}{dx}+{{\left( 1+x+{{x}^{8}} \right)}^{-1}}\dfrac{d\left( {{x}^{8}} \right)}{dx}\]
\[={{x}^{8}}\dfrac{d\left\{ {{\left( 1+x+{{x}^{8}} \right)}^{-1}} \right\}}{dx}+{{\left( 1+x+{{x}^{8}} \right)}^{-1}}\times \left( 8{{x}^{7}} \right)\] ……………………………………..(3)
Using chain rule and on simplifying equation (3), we get
\[\begin{align}
& \Rightarrow {{x}^{8}}\dfrac{d\left\{ {{\left( 1+x+{{x}^{8}} \right)}^{-1}} \right\}}{d\left( 1+x+{{x}^{8}} \right)}\times \dfrac{d\left( 1+x+{{x}^{8}} \right)}{dx}+\dfrac{1}{\left( 1+x+{{x}^{8}} \right)}\left( 8{{x}^{7}} \right) \\
& \Rightarrow {{x}^{8}}\left( -1 \right){{\left( 1+x+{{x}^{8}} \right)}^{-1-1}}\times \left( 8{{x}^{7}}+1 \right)+\dfrac{1}{\left( 1+x+{{x}^{8}} \right)}\left( 8{{x}^{7}} \right) \\
& \Rightarrow -\dfrac{{{x}^{8}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}\left( 8{{x}^{7}}+1 \right)+\dfrac{1}{\left( 1+x+{{x}^{8}} \right)}\left( 8{{x}^{7}} \right) \\
& \Rightarrow -\dfrac{{{x}^{8}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}-\dfrac{{{x}^{8}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}\left( 8{{x}^{7}} \right)+\dfrac{1}{\left( 1+x+{{x}^{8}} \right)}\left( 8{{x}^{7}} \right) \\
& \Rightarrow -\dfrac{{{x}^{8}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}+\dfrac{\left( 8{{x}^{7}} \right)}{\left( 1+x+{{x}^{8}} \right)}\left\{ -\dfrac{{{x}^{8}}}{\left( 1+x+{{x}^{8}} \right)}+1 \right\} \\
& \Rightarrow -\dfrac{{{x}^{8}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}+\dfrac{\left( 8{{x}^{7}} \right)}{\left( 1+x+{{x}^{8}} \right)}\left\{ \dfrac{1+x}{\left( 1+x+{{x}^{8}} \right)} \right\} \\
& \Rightarrow -\dfrac{{{x}^{8}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}+\dfrac{\left( 8{{x}^{7}} \right)}{\left( 1+x+{{x}^{8}} \right)}+\dfrac{8{{x}^{8}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}} \\
& \Rightarrow -\dfrac{7{{x}^{8}}+8{{x}^{7}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}} \\
\end{align}\]
Therefore, option (A) is the correct option.
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