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If $\int \frac{(x - 1)^{2}}{{(x^{2} + 1)}^{2}} d x = \tan^{- 1} x + g (x) + K$ then
$g (x)$ is equal to

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tan−1⁡(x/2)

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Correct option is B

(x−1)2x2+12=x2−2x+1x2+12=x2+1x2+12−2xx2+12=1x2+1−2xx2+12so ∫(x−1)2x2+12dx=∫dxx2+1−∫2xx2+12dx=tan−1⁡x+1x2+1+KTherefore, g(x)=1x2+1

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If ∫(x−1)2x2+12dx=tan−1⁡x+g(x)+K theng(x) is equal to

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Ready to Test Your Skills?

free study material

If $\int \frac{(x - 1)^{2}}{{(x^{2} + 1)}^{2}} d x = \tan^{- 1} x + g (x) + K$ then
$g (x)$ is equal to