Question

If $\int \frac{dx}{x^4+x^3}=\frac{P}{x^2}+\frac{Q}{x}+\ln \left|\frac{x}{x+1}\right|+c$, then

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

Answer 1

The correct option is C $P=-\frac{1}{2},Q=1$

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

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

$\frac{1}{x^3(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x+1}$

$\frac{1}{x^3(x+1)}=\frac{(A{x}^2+Bx+X)(x+1)+D{x}^3}{x^3(x+1)}$

$1=A{x}^2(x+1)+B\left(x\right)(x+1)+C(x+1)+D{x}^3$

Put $x=0$

$1=C$

Put $x=-1$

$1=-D$

$-1=D$

Put $x=1$

$1=2A+2B+2-1$

$2A+2B=0$

$\Rightarrow A=-B$

Put $x=2$

$1=12A+6B+3-8$

$6=12A+6B$

$6=-12B+6B$

$6=-6B$

$B=-1$

$A=1$

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

$\int \frac{dx}{x^3(x+1)}=lnx+\frac{1}{x}-\frac{1}{2x^2}-ln(x+1)+c$

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

$\therefore P=\frac{-1}{2},Q=1$.