Question

If $\int \frac{dx}{x^4+x^3}=\frac{A}{x^2}+\frac{B}{x}+\ln \mid \frac{x}{x+1}\mid +C$ then,

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

Answer 1

Answer: (C)

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

We have,

$\int \frac{dx}{x^4+x^3}=\frac{A}{x^2}+\frac{B}{x}+\ln \mid \frac{x}{x+1}\mid +C$ $............\left(1\right)$

Let

$I=\int \frac{dx}{x^4+x^3}$

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

$\frac{1}{x^3(x+1)}=\frac{P}{x}+\frac{Q}{x^2}+\frac{R}{x^3}+\frac{S}{x+1}$

$1=P{x}^2(x+1)+Qx(x+1)+R(x+1)+S{x}^3$

$1=P(x^3+x^2)+Q(x^2+x)+R(x+1)+S{x}^3$

On comparing $x^3$ term, we get

$0=P+S\Rightarrow P=-S$

On comparing $x^2$ term, we get

$0=P+Q\Rightarrow P=-Q$

On comparing $x$ term, we get

$0=Q+R\Rightarrow Q=-R$

On comparing $constant$ term, we get

$1=R$

So,

$Q=-1,P=1,S=-1$

Therefore,

$I=\int \frac{1}{x}dx+\int \frac{-1}{x^2}dx+\int \frac{1}{x^3}dx+\int \frac{-1}{x+1}dx$

$I=\int \frac{1}{x}dx-\int \frac{1}{x^2}dx+\int \frac{1}{x^3}dx-\int \frac{1}{x+1}dx$

$I=\ln x+\frac{1}{x}-\frac{1}{2x^2}-\ln (x+1)+C$

$I=-\frac{1}{2x^2}+\frac{1}{x}+\ln \left|\frac{x}{x+1}\right|+C$ $........\left(2\right)$

On comparing both equations $\left(1\right)$ and $\left(2\right)$, we get

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

Hence, this is the answer.