The correct option is A 1/3log | 2x 1/2 ( x+1 ) |+C I= nbsp;∫12x^2+x 1dx = 12∫1x^2+x2 12dx = 12∫1 ( x+14 )^2 ( 34 )^2 dx = 12.12 ( ...

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How to Find the Inverse of a Function

∫1/2x2+x-1 dx...

Question

$\int \frac{1}{2 x^{2} + x - 1} d x$ is

\frac{1}{3} log ∣ ∣ ∣ \frac{2 x - 1}{2 (x + 1)} ∣ ∣ ∣ + C

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\frac{1}{3} log ∣ ∣ ∣ \frac{2 x + 1}{2 (x + 1)} ∣ ∣ ∣ + C

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\frac{1}{3} log ∣ ∣ ∣ \frac{2 x - 1}{2 (2 x + 1)} ∣ ∣ ∣ + C

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\frac{1}{3} log ∣ ∣ ∣ \frac{x - 1}{2 (x + 2)} ∣ ∣ ∣ + C

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Solution

The correct option is A $\frac{1}{3} log ∣ ∣ ∣ \frac{2 x - 1}{2 (x + 1)} ∣ ∣ ∣ + C$
$I = \int \frac{1}{2 x^{2} + x - 1} d x$
$= \frac{1}{2} \int \frac{1}{x^{2} + \frac{x}{2} - \frac{1}{2}} d x$
$= \frac{1}{2} \int \frac{1}{{(x + \frac{1}{4})}^{2} - {(\frac{3}{4})}^{2}} d x$
$= \frac{1}{2} . \frac{1}{2 (\frac{3}{4})} log ∣ ∣ ∣ ∣ ∣ ∣ \frac{x + \frac{1}{4} - \frac{3}{4}}{x + \frac{1}{4} + \frac{3}{4}} ∣ ∣ ∣ ∣ ∣ ∣ + C$
$= \frac{1}{3} log ∣ ∣ ∣ ∣ ∣ \frac{x - \frac{1}{2}}{x + 1} ∣ ∣ ∣ ∣ ∣ + C = \frac{1}{3} log ∣ ∣ ∣ \frac{2 x - 1}{2 (x + 1)} ∣ ∣ ∣ + C$

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