Definite Integration
Integration by Parts — Logarithm
nta_pyq_2024_apr
Grade 12
Question:
The value of the integral $\int_{-1}^{2}\log_e\left(x+\sqrt{x^2+1}\right)dx$ is:
$\sqrt{5}-\sqrt{2}+\log_e\left(\dfrac{7+4\sqrt{5}}{1+\sqrt{2}}\right)$
$\sqrt{5}-\sqrt{2}+\log_e\left(\dfrac{9+4\sqrt{5}}{1+\sqrt{2}}\right)$
$\sqrt{2}-\sqrt{5}+\log_e\left(\dfrac{7+4\sqrt{5}}{1+\sqrt{2}}\right)$
$\sqrt{2}-\sqrt{5}+\log_e\left(\dfrac{9+4\sqrt{5}}{1+\sqrt{2}}\right)$
Step-by-Step Solution
Key Concept: Integrate by parts: $I=[x\ln(x+\sqrt{x^2+1})]_{-1}^2-\int_{-1}^2\frac{x}{\sqrt{x^2+1}}dx$. Second integral $=[\sqrt{x^2+1}]_{-1}^2=\sqrt{5}-\sqrt{2}$.
$I=\sqrt{2}-\sqrt{5}+\log_e\left(\dfrac{9+4\sqrt{5}}{1+\sqrt{2}}\right)$.
Correct Answer: 4