Question

If $\int \frac{(2x+3)dx}{x(x+1)(x+2)(x+3)+1}=C-\frac{1}{f\left(x\right)}$, where $f\left(x\right)$ is of the form of $a{x}^2+bx+c$, then

$(a+b+c)$ equals to $\dots \dots$.

• A. $5$
• B. $4$
• C. $3$
• D. $2$

Answer 1

The correct option is A $5$

Let $I=\int \frac{(2x+3)dx}{x(x+1)(x+2)(x+3)+1}$
$=\int \frac{(2x+3)dx}{(x^2+3x)(x^2+3x+2)+1}$
Put $x^2+3x=t$
$\Rightarrow (2x+3)dx=dt$
$I=\int \frac{dt}{t(t+2)+1}$
$=\int \frac{dt}{{(t+1)}^2}=C-\frac{1}{t+1}=C-\frac{1}{x^2+3x+1}$
On comparing with $C-\frac{1}{a{x}^2+bx+c}$, we get
$a=1$, $b=3$, $c=1$
$\Rightarrow a+b+c=5$