CBSE Class 12 Mathematics 2026 Main Paper 65/3/2 · Q303 Markssubstitution
CBSE Class 12 Mathematics 2026 Main Paper 65/3/2 · Question 30 (Integrals)
Evaluate \(\int\frac{dx}{x^{1/2}+x^{1/3}}\). OR Evaluate \(\int\tan^{-1}\!\left(\frac{1-x}{1+x}\right)dx\).
Alternative 1 3 Marks
Evaluate \(\int\frac{dx}{x^{1/2}+x^{1/3}}\).
Answer: 2√x−3x^(1/3)+6x^(1/6)−6 ln(1+x^(1/6))+C, x>0.
- Set t=x^(1/6), so x=t⁶ and dx=6t⁵dt. The integral becomes 6∫t³/(1+t)dt.
- Divide to get t³/(1+t)=t²−t+1−1/(1+t). Integrate term by term.
- Substitute t=x^(1/6) to obtain the stated expression.
Alternative 2 3 Marks
Evaluate \(\int\tan^{-1}\!\left(\frac{1-x}{1+x}\right)dx\).
Answer: πx/4−x arctan x+(1/2)ln(1+x²)+C on x>−1. Replace πx/4 by −3πx/4 on x<−1.
- The tangent subtraction identity gives tan(π/4−arctan x)=(1−x)/(1+x). This angle is principal for x>−1. For x<−1 subtract π to reach the principal range.
- Use integration by parts on ∫arctan x dx, obtaining x arctan x−(1/2)ln(1+x²).
- Integrate the branch constant π/4 or −3π/4 to get the stated antiderivatives on the two components.