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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 \(5(c+d+a b)=\)

  1. A 3
  2. B 5
  3. C 8
  4. D 10
Verified Solution

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…