Question

If $\int \frac{dx}{x^4+x^3}=\frac{A}{x^2}+\frac{B}{x}+\ell n|\frac{x}{x+1}|+C$, 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

The correct option is C $A=-\frac{1}{2},B=1$

Given,

$\int \frac{1}{x^4+x^3}dx$

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

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

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

$=-\frac{1}{2x^2}+\frac{1}{x}-\ln |x+1|+\ln \left|x\right|$

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

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