∫(x dx)/(x^2+3x+2) is equal to:(where c is an arbitrary constant)

The correct answer is Option (1) → \( -\log|x+1|+2\log|x+2|+c \)

Since \( x^2+3x+2=(x+1)(x+2) \), the integrand splits into partial fractions:

    \( \displaystyle \frac{x}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2} \)

    \( x=A(x+2)+B(x+1) \)

    \( x=-1:\ -1=A(1)\ \Rightarrow\ A=-1 \)

    \( x=-2:\ -2=B(-1)\ \Rightarrow\ B=2 \)

So

    \( \displaystyle \int\frac{x\,dx}{x^2+3x+2}=\int\left(\frac{-1}{x+1}+\frac{2}{x+2}\right)dx \)

    \( \displaystyle =-\log|x+1|+2\log|x+2|+c \)

Shortcut: Differentiating Option (1) gives back the integrand:

    \( \displaystyle -\frac{1}{x+1}+\frac{2}{x+2}=\frac{-(x+2)+2(x+1)}{(x+1)(x+2)} \)

    \( \displaystyle =\frac{x}{x^2+3x+2} \)

NCERT (2026-27 edition):
Integration by Partial Fractions