AP EAMCET · Maths · Indefinite Integration
If \(\int \frac{x^4+1}{x^2+1} d x=A x^3+B x^2+C x+D \operatorname{Tan}^{-1} x+E\), then \(A+B+C+D=\)
- A \(\frac{3}{2}\)
- B \(\frac{4}{3}\)
- C \(\frac{1}{3}\)
- D \(\frac{2}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{4}{3}\)
Step-by-step Solution
Detailed explanation
\(\frac{x^4+1}{x^2+1} = x^2-1+\frac{2}{x^2+1}\) \(\int \left(x^2-1+\frac{2}{x^2+1}\right) d x = \frac{x^3}{3}-x+2\operatorname{Tan}^{-1} x+E\) \(A=\frac{1}{3}, B=0, C=-1, D=2\) \(A+B+C+D = \frac{1}{3}+0+(-1)+2 = \frac{4}{3}\)
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