TS EAMCET · Maths · Indefinite Integration
Given that \(\int \frac{1}{x^2+a^2} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c\). If \(\int \frac{1}{x^4+3 x^2+1} d x=a \cdot \tan ^{-1}\left(\frac{b\left(x^2-1\right)}{x}\right)\) \[ +c \tan ^{-1}\left(\frac{d\left(x^2+1\right)}{x}\right)+k \] where \(k\) is a constant of integration, then \(5(c+d+a b)=\)
- A 3
- B 5
- C 8
- D 10
Answer & Solution
Correct Answer
(A) 3
Step-by-step Solution
Detailed explanation
we have \(\int \frac{1}{x^4+3 x^2+1} d x\) now dividing by \(\mathrm{x}^2\) to the numerator and denominator…
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