Question

If $\int \frac{dx}{x^4+x^3}=\frac{A}{x^2}+\frac{B}{x}+\log \left|\frac{x}{x+1}\right|+c$, then

• A. $A=\frac{1}{2},B=1$.
• B. $A=1,B=\frac{1}{2}$.
• C. $A=-\frac{1}{2},B=1$.
• D. $A=1,B=1$.

Answer 1

The correct option is C

$A=-\frac{1}{2},B=1$.

Explanation for the correct option:

Step-1: Solve the partial fraction:

The given equation, $\int \frac{dx}{x^4+x^3}=\frac{A}{x^2}+\frac{B}{x}+\log \left|\frac{x}{x+1}\right|+c$.

Consider the expression, $\frac{1}{x^4+x^3}=\frac{1}{x^3\left(x+1\right)}$

$\begin{array}{crcl}\Rightarrow & \frac{1}{x^4+x^3}& =& \frac{P}{x+1}+\frac{Q{x}^2+Rx+S}{x^3}\\ \Rightarrow & \frac{1}{x^4+x^3}& =& \frac{P{x}^3+\left(Q{x}^2+Rx+S\right)\left(x+1\right)}{x^3\left(x+1\right)}\\ \Rightarrow & \frac{1}{x^4+x^3}& =& \frac{P{x}^3+Q{x}^3+R{x}^2+Sx+Q{x}^2+Rx+S}{x^4+x^3}\\ \Rightarrow & \left(P+Q\right)x^3+\left(R+Q\right)x^2+\left(S+R\right)x+S& =& 1\\ \Rightarrow & \left(P+Q\right)x^3+\left(R+Q\right)x^2+\left(S+R\right)x+S& =& 0·x^3+0·x^2+0·x+1\end{array}$

Thus, compare both sides of the equation.

$P+Q=0;R+Q=0;S+R=0;S=1;$

From the above equations, the values can be given by, $P=-1,Q=1,R=-1,S=1$.

Thus, $\frac{1}{x^4+x^3}=\frac{-1}{x+1}+\frac{x^2-x+1}{x^3}$.

$\Rightarrow \frac{1}{x^4+x^3}=-\frac{1}{x+1}+\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3}$

Step-2: Integrate the given partial fraction:

$\int \frac{dx}{x^4+x^3}=\int \left(-\frac{1}{x+1}+\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3}\right)dx$

$$\begin{gathered} =-\int \frac{dx}{x+1}+\int \frac{dx}{x}-\int \frac{dx}{x^2}+\int \frac{dx}{x^3} \\ =-\log \left|x+1\right|+\log \left|x\right|-\frac{x^{-2+1}}{\left(-2+1\right)}+\frac{x^{-3+1}}{\left(-3+1\right)}+c \\ =-\log \left|x+1\right|+\log \left|x\right|-\frac{x^{-1}}{-1}+\frac{x^{-2}}{-2}+c \\ =\frac{-\frac{1}{2}}{x^2}+\frac{1}{x}+\log \left|\frac{x}{x+1}\right|+c\mathbf{}\left[\because \log \left(m\right)-\log \left(n\right)=\log \left(\frac{m}{n}\right)\right] \end{gathered}$$

Compare the equations $\int \frac{dx}{x^4+x^3}=\frac{A}{x^2}+\frac{B}{x}+\log \left|\frac{x}{x+1}\right|+c$ and $\int \frac{dx}{x^4+x^3}=\frac{-{\displaystyle \frac{1}{2}}}{x^2}+\frac{1}{x}+\log \left|\frac{x}{x+1}\right|+c$.

Therefore, $A=-\frac{1}{2},B=1$.

Hence, option(C) is the correct option.