∫x+tan−1⁡3x21+9x2dx=Alog⁡1+9x2+Btan−1⁡3x3+c then BA=

I=∫x+tan−1⁡(3x)21+9x2dx=∫xdx1+9x2+∫tan−1⁡3x21+9x2dx=118∫18x1+9x2dx+13∫tan−1⁡3x21+9x2dx Let 1+9x2=t&tan−1⁡3x=Z18xdx=dt&31+9x2dx=dzI=118∫t1 dt+13∫(z)2dz=118log⁡1+9x2+19tan−1⁡(3x)3+cA=118,B=19⇒BA=2

∫x+tan−1⁡3x21+9x2dx=Alog⁡1+9x2+Btan−1⁡3x3+c then BA=

I=∫x+tan−1⁡(3x)21+9x2dx=∫xdx1+9x2+∫tan−1⁡3x21+9x2dx=118∫18x1+9x2dx+13∫tan−1⁡3x21+9x2dx Let 1+9x2=t&tan−1⁡3x=Z18xdx=dt&31+9x2dx=dzI=118∫t1 dt+13∫(z)2dz=118log⁡1+9x2+19tan−1⁡(3x)3+cA=118,B=19⇒BA=2

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$\int \frac{x + {(\tan^{- 1} 3 x)}^{2}}{1 + 9 x^{2}} d x = A \log (1 + 9 x^{2}) + B {(\tan^{- 1} 3 x)}^{3} + c then \frac{B}{A} =$

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$\frac{2}{3}$

$\frac{3}{2}$

answer is B.

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Detailed Solution

$\begin{array}{l} I = \int \frac{x + {(\tan^{- 1} (3 x))}^{2}}{1 + 9 x^{2}} d x = \int \frac{x d x}{1 + 9 x^{2}} + \int \frac{{(\tan^{- 1} 3 x)}^{2}}{1 + 9 x^{2}} d x \\ = \frac{1}{18} \int \frac{18 x}{1 + 9 x^{2}} d x + \frac{1}{3} \int \frac{{(\tan^{- 1} 3 x)}^{2}}{1 + 9 x^{2}} d x \\ Let 1 + 9 x^{2} = t & \tan^{- 1} 3 x = Z \\ 18 x d x = d t & \frac{3}{1 + 9 x^{2}} d x = d z \\ I = \frac{1}{18} \int_{t}^{1} d t + \frac{1}{3} \int (z)^{2} d z \\ = \frac{1}{18} \log (1 + 9 x^{2}) + \frac{1}{9} {(\tan^{- 1} (3 x))}^{3} + c \\ A = \frac{1}{18}, B = \frac{1}{9} \Rightarrow \frac{B}{A} = 2 \end{array}$