Question

If $\int \log (x^2+x)dx=x\log x+(x+1)\log (x+1)+K,$ then K is equal to

• A. $2x+\log (x+1)+C$
• B. $2x-\log (x+1)+C$
• C. constant
• D. none of these

Answer 1

The correct option is D none of these

$\int \log (x^2+x)dx=x\log (x^2+x)-\int \frac{x(2x+1)}{x^2+x}dx+c$
$=x\log (x^2+x)-\int \frac{(2x+1)}{x^2+x}dx+c=x\log (x^2+x)-\int (2-\frac{1}{x+1})dx+c$
$=x\log (x^2+x)-2x+\log (x+1)+c=x\log x+(x+1)\log (x+1)-2x+c\therefore k=-2x+c$