Indefinite Integration
Substitution — Inverse Trig in Denominator
nta_pyq_2024_jan
Grade 12
Question:
The integral $\displaystyle\int\dfrac{(x^8-x^2)\,dx}{(x^{12}+3x^6+1)\tan^{-1}\!\left(x^3+\dfrac{1}{x^3}\right)}$ is equal to:
$\log_e\left(\left|\tan^{-1}\!\left(x^3+\dfrac{1}{x^3}\right)\right|\right)^{1/3}+C$
$\log_e\left(\left|\tan^{-1}\!\left(x^3+\dfrac{1}{x^3}\right)\right|\right)^{1/2}+C$
$\log_e\left|\tan^{-1}\!\left(x^3+\dfrac{1}{x^3}\right)\right|+C$
$\log_e\left(\left|\tan^{-1}\!\left(x^3+\dfrac{1}{x^3}\right)\right|\right)^{3}+C$
Step-by-Step Solution
Key Concept: Let $t=\tan^{-1}\!\left(x^3+\frac{1}{x^3}\right)$. Differentiate: $dt=\frac{1}{1+(x^3+x^{-3})^2}\cdot(3x^2-3x^{-4})dx$. Show the numerator $x^8-x^2$ relates to $dt$ with a factor, reducing the integral to $\frac{1}{3}\int\frac{dt}{t}$.
Let $t=\tan^{-1}(x^3+x^{-3})$. $dt=\frac{3x^2-3x^{-4}}{1+(x^3+x^{-3})^2}dx=\frac{3(x^8-x^2)/x^4}{(x^{12}+3x^6+1)/x^6}dx=\frac{3(x^8-x^2)}{x^{12}+3x^6+1}dx$. $I=\frac{1}{3}\int\frac{dt}{t}=\frac{\ln|t|}{3}+C=\ln\left|\tan^{-1}\!\left(x^3+\frac{1}{x^3}\right)\right|^{1/3}+C$.
Correct Answer: 1