Indefinite Integration
Indefinite Integration
nta_abhyas_2025
Grade 12
Question:
$\int \frac{dx}{x(x+1)}$ is equal to (where $C$ is an arbitrary constant)
$\ln\left|\frac{x+1}{x}\right| + C$
$\frac{1}{2}\ln\left|\frac{x+1}{x}\right|^2 + C$
$\ln\left|\frac{x}{x+1}\right| + C$
$\ln\left|\frac{x+1}{x}\right|^2 + C$
Step-by-Step Solution
Key Concept: Use substitution with logarithmic expressions to transform the integral into a simpler polynomial form
Put $\ln(x+1) - \ln x = t$, which gives $\frac{1}{x+1} - \frac{1}{x}dx = dt$, or equivalently $\frac{dt}{dx} = \frac{-1}{x(x+1)}$. Therefore $\frac{dx}{x(x+1)} = -dt$. The integral becomes $-\int t dt = -\frac{t^2}{2} + C = -\frac{1}{2}\left[\ln\left(\frac{x+1}{x}\right)\right]^2 + C$.
Correct Answer: 1