Question

$\int \frac{1}{x^3(x^3-1)}dx=$

• A. $\frac{1}{3}\log \left|\frac{x^3}{x^3-1}\right|+C$
• B. $\left(\frac{1}{3}\right)\log \left|\frac{1-x^3}{x^3}\right|+C$
• C. $\log \left|\frac{x^3}{x^3-1}\right|+C$
• D. $\left(\frac{1}{3}\right)\log \left|x-1\right|-\left(\frac{1}{6}\right)\log \left|(x^2+x+1)\right|-\left(\frac{1}{\sqrt{3}}\right){\tan }^{-1}\left(\frac{\left(x+\frac{1}{2}\right)}{\frac{\sqrt{3}}{2}}\right)+C$

Answer 1

The correct option is D

$\left(\frac{1}{3}\right)\log \left|x-1\right|-\left(\frac{1}{6}\right)\log \left|(x^2+x+1)\right|-\left(\frac{1}{\sqrt{3}}\right){\tan }^{-1}\left(\frac{\left(x+\frac{1}{2}\right)}{\frac{\sqrt{3}}{2}}\right)+C$

Explanation for the correct option:

Step 1: Expanding the given integral into two parts and defining the integral $I_1$

$$\begin{gathered} \int \frac{1}{x^3(x^3-1)}dx=\int \frac{1+x^3-x^3}{x^3(x^3-1)}dx\left[\text{add and subtract}{x}^3\text{in numerator}\right] \\ =\int \frac{x^3-\left(x^3-1\right)}{x^3(x^3-1)}dx \\ =\int \frac{x^3}{x^3(x^3-1)}dx-\int \frac{\left(x^3-1\right)}{x^3(x^3-1)}dx \\ =\int \frac{1}{(x^3-1)}dx-\int \frac{1}{x^3}dx \\ ={\text{I}}_1-{\text{I}}_2 \end{gathered}$$

Let

$$\begin{gathered} {\text{I}}_1\text{=}\int \frac{1}{(x^3-1)}dx \\ =\int \frac{1}{(x-1)(x^2+x+1)}dx...\left(i\right) \end{gathered}$$

Step 2: Finding the value of $A,B,\mathrm{and}C$

To evaluate this integral using the integration of rational numbers

$$\begin{gathered} \frac{1}{(x-1)(x^2+x+1)}=\frac{A}{(x-1)}+\frac{Bx+C}{(x^2+x+1)}...\left(ii\right) \\ \Rightarrow \frac{1}{(x-1)(x^2+x+1)}=\frac{A(x^2+x+1)+\left(Bx+C\right)(x-1)}{(x-1)(x^2+x+1)} \\ \Rightarrow 1=A(x^2+x+1)+\left(Bx+C\right)(x-1)...\left(iii\right) \end{gathered}$$

Substitute $x=1\text{in equation}\left(iii\right)$

$$\begin{gathered} \Rightarrow 1=A(1+1+1)+(Bx+c)(1-1) \\ \Rightarrow 1=3A \\ \Rightarrow A=\frac{1}{3} \end{gathered}$$

Substitute $x=0\text{in equation}\left(iii\right)$

$\begin{array}{rcl}& \Rightarrow & 1=A(0+0+1)+(B\left(0\right)+c)(0-1)\\ & \Rightarrow & 1=A-C\\ & \Rightarrow & 1=\frac{1}{3}-C\left[\because A=\frac{1}{3}\right]\\ & \Rightarrow & C=\frac{-2}{3}\end{array}$

Substitute $x=-1\text{in equation}\left(iii\right)$

$\begin{array}{rcl}& \Rightarrow & 1=A(1-1+1)+(B\left(-1\right)+c)(-1-1)\\ & \Rightarrow & 1=A+2B-2C\\ & \Rightarrow & 1=\frac{1}{3}+2B-2\left(\frac{-2}{3}\right)\left[\because A=\frac{1}{3},C=\frac{-2}{3}\right]\\ & \Rightarrow & 1=\frac{5}{3}+2B\\ & \Rightarrow & B=\frac{1}{2}\left(1-\frac{5}{3}\right)\\ & \Rightarrow & B=\frac{1}{2}\left(-\frac{2}{3}\right)\\ & \Rightarrow & B=-\frac{1}{3}\end{array}$

Substituting the value of $A,B,C\text{in equation}\left(ii\right)$

$\begin{array}{rcl}\frac{1}{(x-1)(x^2+x+1)}& =& \frac{\frac{1}{3}}{(x-1)}+\frac{\left(\frac{-1}{3}x+\left(-\frac{2}{3}\right)\right)}{(x^2+x+1)}\left[\because A=\frac{1}{3},B=-\frac{1}{3},C=\frac{-2}{3}\right]\\ & \Rightarrow & \frac{1}{(x-1)(x^2+x+1)}=\frac{1}{3}\frac{1}{(x-1)}-\frac{1}{3}\frac{\left(x+2\right)}{(x^2+x+1)}\end{array}$

Step 3: Integrating ${\text{I}}_1$

Substitute this in ${\text{I}}_1$

$$\begin{gathered} \Rightarrow {\text{I}}_1=\int \frac{1}{(x-1)(x^2+x+1)}dx \\ =\int \frac{1}{3}\frac{1}{(x-1)}-\frac{1}{3}\frac{\left(x+2\right)}{(x^2+x+1)}dx \\ =\frac{1}{3}\left[\int \frac{1}{(x-1)}dx-\int \frac{\left(x+2\right)}{(x^2+x+1)}dx\right] \\ =\frac{1}{3}\left[\int \frac{1}{(x-1)}dx-\int \frac{2\left(x+2\right)}{2(x^2+x+1)}dx\right]\left[\text{multiply and divide by}2\text{in second term}\right] \\ =\frac{1}{3}\left[\int \frac{1}{(x-1)}dx-\frac{1}{2}\int \frac{\left(2x+4\right)}{(x^2+x+1)}dx\right] \\ =\frac{1}{3}\left[\int \frac{1}{(x-1)}dx-\frac{1}{2}\int \frac{\left(2x+1\right)+3}{(x^2+x+1)}dx\right] \\ =\frac{1}{3}\left[\int \frac{1}{(x-1)}dx-\frac{1}{2}\int \left(\frac{\left(2x+1\right)}{(x^2+x+1)}+\frac{3}{(x^2+x+1)}\right)dx\right] \\ =\frac{1}{3}\left[\int \frac{1}{(x-1)}dx-\frac{1}{2}\int \frac{\left(2x+1\right)}{(x^2+x+1)}dx-\frac{3}{2}\int \frac{1}{(x^2+x+1)}dx\right] \\ =\frac{1}{3}\left[\int \frac{1}{(x-1)}dx-\frac{1}{2}\int \frac{\left(2x+1\right)}{(x^2+x+1)}dx-\frac{3}{2}\int \frac{1}{{\left(x+{\displaystyle \frac{1}{2}}\right)}^2+{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2}dx\right]...\left(iv\right)\left[\because x^2+x+1=x^2+2\frac{1}{2}x+\frac{1}{2}+\frac{3}{2}\right] \end{gathered}$$Step 4: Evaluating the integral using the substitution method

Evaluating $\int \frac{1}{(x-1)}dx$

$\int \frac{1}{(x-1)}dx=\log \left|x-1\right|\left[\because \int \frac{1}{x}dx=log\left|x\right|\right]$

Evaluating $\int \frac{\left(2x+1\right)}{(x^2+x+1)}dx$

$$\begin{gathered} \int \frac{\left(2x+1\right)}{(x^2+x+1)}dx=\int \frac{dt}{t}\mathbf{}\left[\because t=(x^2+x+1)\Rightarrow dt=\left(2x+1\right)\right] \\ =\log \left|t\right| \\ =\log \left|(x^2+x+1)\right| \end{gathered}$$

Evaluating $\int \frac{1}{{\left(x+{\displaystyle \frac{1}{2}}\right)}^2+{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2}dx$

$$\begin{gathered} \int \frac{1}{{\left(x+{\displaystyle \frac{1}{2}}\right)}^2+{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2}dx=\frac{1}{\frac{\sqrt{3}}{2}}{\tan }^{-1}\left(\frac{\left(x+\frac{1}{2}\right)}{\frac{\sqrt{3}}{2}}\right)\left[\because \int \frac{1}{x^2+a^2}dx=\frac{1}{a}ta{n}^{-1}\left(\frac{x}{a}\right)\right] \\ =\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{\left(x+\frac{1}{2}\right)}{\frac{\sqrt{3}}{2}}\right) \end{gathered}$$

Step 5: Finding the value of $\int \frac{1}{x^3(x^3-1)}dx$

Substituting the obtained values in equation $\left(iv\right)$ we get,

$$\begin{gathered} =\frac{1}{3}\left[\log \left|x-1\right|-\frac{1}{2}\log \left|(x^2+x+1)\right|-\frac{3}{2}\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{\left(x+\frac{1}{2}\right)}{\frac{\sqrt{3}}{2}}\right)\right] \\ =\left(\frac{1}{3}\right)\log \left|x-1\right|-\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)\log \left|(x^2+x+1)\right|-\left(\frac{1}{3}\right)\left(\frac{3}{2}\right)\left(\frac{2}{\sqrt{3}}\right){\tan }^{-1}\left(\frac{\left(x+\frac{1}{2}\right)}{\frac{\sqrt{3}}{2}}\right) \\ =\left(\frac{1}{3}\right)\log \left|x-1\right|-\left(\frac{1}{6}\right)\log \left|(x^2+x+1)\right|-\left(\frac{1}{\sqrt{3}}\right){\tan }^{-1}\left(\frac{\left(x+\frac{1}{2}\right)}{\frac{\sqrt{3}}{2}}\right)+C \end{gathered}$$

Hence, option (D) is the correct answer.