CBSE Class 12 Mathematics 2025: question 32(a)

Hints

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Step-by-step solution

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1. Put $x^2=t$ (for the partial fractions only): $\frac{t+1}{(t+2)(2t+1)}=\frac{A}{t+2}+\frac{B}{2t+1}$, so $t+1=A(2t+1)+B(t+2)$.

2. At $t=-2$: $-1=-3A$, so $A=\frac{1}{3}$. At $t=-\frac{1}{2}$: $\frac{1}{2}=\frac{3}{2}B$, so $B=\frac{1}{3}$.

   2 marks

3. $\int \frac{x^2+1}{(x^2+2)(2x^2+1)}dx=\frac{1}{3}\int \frac{dx}{x^2+2}+\frac{1}{3}\int \frac{dx}{2x^2+1}$

   ½ mark

4. $\frac{1}{3}\int \frac{dx}{x^2+(\sqrt{2}{)}^2}=\frac{1}{3}\cdot \frac{1}{\sqrt{2}}{\tan }^{-1}\frac{x}{\sqrt{2}}$

   1 mark

5. $\frac{1}{3}\int \frac{dx}{2x^2+1}=\frac{1}{6}\int \frac{dx}{x^2+(1/\sqrt{2}{)}^2}=\frac{1}{6}\cdot \sqrt{2}{\tan }^{-1}(\sqrt{2}x)=\frac{1}{3\sqrt{2}}{\tan }^{-1}(\sqrt{2}x)$

   1 mark

6. $\int \frac{x^2+1}{(x^2+2)(2x^2+1)}dx=\frac{1}{3\sqrt{2}}\left[{\tan }^{-1}\frac{x}{\sqrt{2}}+{\tan }^{-1}(\sqrt{2}x)\right]+C$

   ½ mark

Answer: $\frac{1}{3\sqrt{2}}\left[{\tan }^{-1}\frac{x}{\sqrt{2}}+{\tan }^{-1}(\sqrt{2}x)\right]+C$

Revise the chapter

• Chapter: Integrals (notes, formulas, where marks are lost and every past question)
• Topic: Integration by Partial Fractions