Integral Calculus
Integration by Substitution
MMTS_Full_Test_21
Grade 12
Question:
$\displaystyle\int\frac{6x^{10}+4}{x^3\sqrt{x^{10}-3x^4-1}}\,dx,\;x>0$
$\dfrac{\sqrt{x^{10}-3x^4-1}}{x^2}+c$
$\dfrac{2\sqrt{x^{10}-3x^4-1}}{x^2}+c$
$2\sqrt{x^{10}-3x^4-1}+c$
$\sqrt{x^{10}-3x^4}+C$
Step-by-Step Solution
Key Concept: Substitute $t=x^{10}-3x^4-1$
Let $t=x^{10}-3x^4-1$; $dt=(10x^9-12x^3)dx=2x^3(5x^6-6)dx$. Simplify to get $\frac{2\sqrt{t}}{x^2}+c$.
Correct Answer: 2