Let fx=2x+1 and gx=∫fx/(x^2x+1^2)dx. If 6g2+1=0, then g-1/2 is equal to

Correct answer: (A) 4. gx=∫(x+1^2-x^2)/(x^2x+1^2)dx=∫(1x^2-1x+1^2dx=-1)/x+1/(x+1)+c =-1/(xx+1)+c g2=-1/6+c=-1/6⇒c=0 Hence, g-1/2=4

Q13 / 104 · NTA Abhyas (JEE Main) 2020 · Mathematics · Indefinite Integration

Let $f (x) = 2 x + 1$ and $g (x) = \int \frac{f (x)}{x^{2} {(x + 1)}^{2}} d x$ . If $6 g (2) + 1 = 0,$ then $g (- \frac{1}{2})$ is equal to

g (x) = \int \frac{{(x + 1)}^{2} - x^{2}}{x^{2} {(x + 1)}^{2}} d x = \int (\frac{1}{x^{2}} - \frac{1}{{(x + 1)}^{2}}) d x = - \frac{1}{x} + \frac{1}{x + 1} + c

= - \frac{1}{x (x + 1)} + c

g (2) = - \frac{1}{6} + c = - \frac{1}{6} \Rightarrow c = 0

Hence,

g (- \frac{1}{2}) = 4

Hence, the correct answer is option: 'A'.