If ft+12t+1=t+1 then ∫f(x)dx=

ft+12t+1=t+1 t+12t+1=x⇒t+1=2xt+x ⇒1-x=t(2x-1) ⇒t=1-x2x-1 ft+12t+1=t+1 f(x)=1-x2x-1+1=x2x-1 f(x)=121+12x-1∫f(x)dx=12∫1+12x-1dx =x2+14log2x-1+C

If ft+12t+1=t+1 then ∫f(x)dx=

ft+12t+1=t+1 t+12t+1=x⇒t+1=2xt+x ⇒1-x=t(2x-1) ⇒t=1-x2x-1 ft+12t+1=t+1 f(x)=1-x2x-1+1=x2x-1 f(x)=121+12x-1∫f(x)dx=12∫1+12x-1dx =x2+14log2x-1+C

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If $f (\frac{t + 1}{2 t + 1}) = t + 1$ then $\int f (x) d x =$

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$\frac{x^{2}}{2} + C$

$\log (2 x - 1) + \frac{1}{2} \log (x + 1) + C$

$\frac{1}{2} \log (2 x - 1) + C$

$\frac{x}{2} + \frac{1}{4} \log |2 x - 1| + C$

answer is D.

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Detailed Solution

$f (\frac{t + 1}{2 t + 1}) = t + 1 \frac{t + 1}{2 t + 1} = x \Rightarrow t + 1 = 2 x t + x \Rightarrow 1 - x = t (2 x - 1) \Rightarrow t = \frac{1 - x}{2 x - 1} f (\frac{t + 1}{2 t + 1}) = t + 1 f (x) = \frac{1 - x}{2 x - 1} + 1 = \frac{x}{2 x - 1} f (x) = \frac{1}{2} (1 + \frac{1}{2 x - 1})$
$\int f (x) d x = \frac{1}{2} \int (1 + \frac{1}{2 x - 1}) d x = \frac{x}{2} + \frac{1}{4} \log |2 x - 1| + C$