CBSE Class 12 Mathematics 2023: question 28(b)

Hints

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Step-by-step solution

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1. The denominator is $(x-1)(x^2+1)=x^3-x^2+x-1$.

   By division, $x^4=(x+1)(x^3-x^2+x-1)+1$. So

   $\frac{x^4}{(x-1)(x^2+1)}=x+1+\frac{1}{(x-1)(x^2+1)}$

   1 mark

2. Let $\frac{1}{(x-1)(x^2+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}$. Then $1=A(x^2+1)+(Bx+C)(x-1)$.

   • Putting $x=1$: $A=\frac{1}{2}$.
   • Comparing $x^2$ terms: $A+B=0$, so $B=-\frac{1}{2}$.
   • Comparing constants: $A-C=1$, so $C=-\frac{1}{2}$.

   So $\frac{1}{(x-1)(x^2+1)}=\frac{1}{2(x-1)}-\frac{1}{2}\cdot \frac{x+1}{x^2+1}$

   1 mark

3. $I=\int \left[x+1+\frac{1}{2(x-1)}-\frac{1}{4}\cdot \frac{2x}{x^2+1}-\frac{1}{2}\cdot \frac{1}{x^2+1}\right]dx$

   ½ mark

4. $=\frac{x^2}{2}+x+\frac{1}{2}\log |x-1|-\frac{1}{4}\log (x^2+1)-\frac{1}{2}{\tan }^{-1}x+C$

   ½ mark

Answer: $\frac{x^2}{2}+x+\frac{1}{2}\log |x-1|-\frac{1}{4}\log (x^2+1)-\frac{1}{2}{\tan }^{-1}x+C$

Revise the chapter

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