If $\frac{x^2+1}{\left(x^2+2\right)\left(x^2+3\right)}=\frac{A x+B}{x^2+2}+\frac{C x+D}{x^2+3}$, then $A+B+C+D=$
- A
0
- B
1
- C
-1
- D
6
✓ Correct answer: B
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NewNcert › mathematics › Indefinite Integration
If $\frac{x^2+1}{\left(x^2+2\right)\left(x^2+3\right)}=\frac{A x+B}{x^2+2}+\frac{C x+D}{x^2+3}$, then $A+B+C+D=$
0
1
-1
6
Want the full step-by-step reasoning?
Question ID #184272 · NewNcert