Question

Consider the given expression $\int{\dfrac{7{{x}^{8}}+8{{x}^{7}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}dx=f\left( x \right)+c}$ then find $f\left( x \right)$

Answer 1

Hint: Take ${{x}^{8}}$ as common from the bracket of the denominator of the expression i.e. taking common ${{x}^{16}}$ as common from the denominator of the expression. And now suppose the term in bracket ‘t’ and differentiate it w.r.t $x$ . Now get the expression in terms of ‘t’ and get the integration with the help of formula
$\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$

Complete step-by-step answer:
Given expression in the problem is
$\int{\dfrac{7{{x}^{8}}+8{{x}^{7}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}dx=f\left( x \right)+c}.................\left( i \right)$
So, let us suppose the integral given in the right-hand side of the equation (i) is I. So, we get
$I=\int{\dfrac{7{{x}^{8}}+8{{x}^{7}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}dx...................\left( ii \right)}$
Taking ${{x}^{8}}$ common from the bracket of the denominator of the equation (ii). So, we can re-write the equation (ii) as

& \Rightarrow I=\int{\dfrac{7{{x}^{8}}+8{{x}^{7}}}{{{\left( {{x}^{8}}\left( \dfrac{1}{{{x}^{8}}}+\dfrac{x}{{{x}^{8}}}+\dfrac{{{x}^{8}}}{{{x}^{8}}} \right) \right)}^{2}}}}dx \\\ & \Rightarrow I=\int{\dfrac{7{{x}^{8}}+8{{x}^{7}}}{{{\left( {{x}^{8}}\left( {{x}^{-8}}+{{x}^{-7}}+1 \right) \right)}^{2}}}dx} \\\ \end{aligned}$$ We can re-write the above equation as $$\Rightarrow I=\int{\dfrac{7{{x}^{8}}+8{{x}^{7}}}{{{x}^{16}}{{\left( {{x}^{-8}}+{{x}^{-7}}+1 \right)}^{2}}}dx}$$ Now, dividing the numerator by ${{x}^{16}}$ , we get $$\Rightarrow I=\int{\dfrac{7{{x}^{-8}}+8{{x}^{-9}}}{{{\left( {{x}^{-8}}+{{x}^{-7}}+1 \right)}^{2}}}dx}................\left( iii \right)$$ Let us suppose ${{x}^{-8}}+{{x}^{-7}}+1$ as ‘t’. So, we can write an equation as $t={{x}^{-8}}+{{x}^{-7}}+1..............\left( iv \right)$ Differentiating the above equation w.r.t. $'x'$ as $\begin{aligned} & \Rightarrow \dfrac{dt}{dx}=\dfrac{d}{dx}\left( {{x}^{-8}}+{{x}^{-7}}+1 \right) \\\ & \Rightarrow \dfrac{dt}{dx}=\dfrac{d}{dx}{{x}^{-8}}+\dfrac{d}{dx}{{x}^{-7}}+\dfrac{d}{dx}\left( 1 \right) \\\ \end{aligned}$ We know $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}},\dfrac{d}{dx}\left( \text{constant} \right)=0$ So, we can write $\dfrac{dt}{dx}$ as $\begin{aligned} & \Rightarrow \dfrac{dt}{dx}=-8{{x}^{-8-1}}+\left( -7 \right){{x}^{-7-1}}+0 \\\ & \Rightarrow \dfrac{dt}{dx}=-8{{x}^{-9}}-7{{x}^{-8}} \\\ & \Rightarrow dt=-\left( 7{{x}^{-8}}+8{{x}^{-9}} \right)dx.................\left( v \right) \\\ \end{aligned}$ Now, we can replace ${{x}^{-8}}+{{x}^{-7}}+1$ by ‘t’ from the equation (iv) and $\left( 7{{x}^{-8}}+8{{x}^{-9}} \right)dx$ as $'-dt'$ from the equation (v). So, we can re-write integral I in terms of ‘t’ as $\begin{aligned} & \Rightarrow I=\int{\dfrac{-dt}{{{t}^{2}}}} \\\ & \Rightarrow I=-\int{{{t}^{-2}}dt}.........................\left( vi \right) \\\ \end{aligned}$ We know the integration of ${{x}^{n}}$ is given as $\Rightarrow \int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}.................\left( vii \right)}$ Hence, we get integral I of equation (vi), with the help of equation (vii) as $\begin{aligned} & \Rightarrow I=-\left( \dfrac{{{t}^{-2+1}}}{-2+1} \right)+c=\dfrac{-{{t}^{-1}}}{-1}+c \\\ & \Rightarrow I=\dfrac{1}{t}+c \\\ \end{aligned}$ Now, we can get integral I in terms of $x$ using equation (vi) as $\begin{aligned} & \Rightarrow I=\dfrac{1}{{{x}^{-8}}+{{x}^{-7}}+1}+c \\\ & \Rightarrow I=\dfrac{1}{\dfrac{1}{{{x}^{8}}}+\dfrac{1}{{{x}^{7}}}+1}+c \\\ & \Rightarrow I=\dfrac{{{x}^{8}}}{1+x+{{x}^{8}}}+c \\\ \end{aligned}$ Now, comparing the above equation, with the R.H.S. of equation (i), we get $f\left( x \right)=\dfrac{{{x}^{8}}}{1+x+{{x}^{8}}}$ Hence, option (a) is the correct answer. Note: Another way of observing the integral would be that we can multiply the numerator and denominator by ${{x}^{16}}$ and hence, we get $$\begin{aligned} & \Rightarrow I=\int{\dfrac{\left( 7{{x}^{8}}+8{{x}^{7}} \right)\times {{x}^{16}}}{{{\left( 1+x+{{x}^{8}} \right)}^{2}}\times {{x}^{16}}}dx} \\\ & \Rightarrow I=\int{\dfrac{\dfrac{7{{x}^{8}}+8{{x}^{7}}}{{{x}^{16}}}}{\dfrac{{{\left( 1+x+{{x}^{8}} \right)}^{2}}}{{{x}^{16}}}}dx} \\\ & \Rightarrow I=\int{\dfrac{7{{x}^{-8}}+8{{x}^{-9}}}{{{\left( \dfrac{1+x+{{x}^{8}}}{{{x}^{8}}} \right)}^{2}}}dx} \\\ & \Rightarrow I=\dfrac{7{{x}^{-8}}+8{{x}^{-9}}}{{{\left( {{x}^{-8}}+{{x}^{-7}}+1 \right)}^{2}}} \\\ \end{aligned}$$ So, it can be another approach to re-write the given integral. Do not expand ${{\left( 1+x+{{x}^{8}} \right)}^{2}}$ to make the given integral complex. Taking out ${{x}^{8}}$ from the bracket of the denominator of the expression is the key point of the problem and there are a number of problems based on this approach. So, do remember this concept for future reference.