Indefinite Integration
Integration by Parts — Logarithmic Result
nta_pyq_2023_apr
Grade 12
Question:
Let $I(x)=\displaystyle\int\dfrac{x^2(x\sec^2x+\tan x)}{(x\tan x+1)^2}\,dx$. If $I(0)=0$, then $I\!\left(\dfrac{\pi}{4}\right)$ is equal to
$\log_e\dfrac{(\pi+4)^2}{16}+\dfrac{\pi^2}{4(\pi+4)}$
$\log_e\dfrac{(\pi+4)^2}{16}-\dfrac{\pi^2}{4(\pi+4)}$
$\log_e\dfrac{(\pi+4)^2}{32}-\dfrac{\pi^2}{4(\pi+4)}$
$\log_e\dfrac{(\pi+4)^2}{32}+\dfrac{\pi^2}{4(\pi+4)}$
Step-by-Step Solution
Key Concept: Integrate by parts: $\int\frac{x^2(x\sec^2x+\tan x)}{(x\tan x+1)^2}dx=-\frac{x^2}{x\tan x+1}+2\int\frac{x\,dx}{x\tan x+1}$. The second integral: $2\ln|x\sin x+\cos x|+C$.
$I(\frac{\pi}{4})=\ln\frac{(\pi+4)^2}{32}-\frac{\pi^2}{4(\pi+4)}$.
Correct Answer: 3