If int frac{(x - 1)^2}{(x^2 + 1)^2} dx = tan^ | vedclass

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(B) Given the integral $I = \int \frac{(x - 1)^2}{(x^2 + 1)^2} dx$. Expanding the numerator,we get $(x - 1)^2 = x^2 - 2x + 1$. So,$I = \int \frac{x^2

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$\frac{1}{x^2 + 1}$

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(B) Given the integral $I = \int \frac{(x - 1)^2}{(x^2 + 1)^2} dx$. Expanding the numerator,we get $(x - 1)^2 = x^2 - 2x + 1$. So,$I = \int \frac{x^2 + 1 - 2x}{(x^2 + 1)^2} dx$. Splitting the integral,we have $I = \int \frac{x^2 + 1}{(x^2 + 1)^2} dx - \int \frac{2x}{(x^2 + 1)^2} dx$. This simplifies to $I = \int \frac{1}{x^2 + 1} dx - \int 2x(x^2 + 1)^{-2} dx$. We know that $\int \frac{1}{x^2 + 1} dx = 	an^{-1}(x)$. For the second part,let $u = x^2 + 1$,then $du = 2x dx$. Thus,$\int 2x(x^2 + 1)^{-2} dx = \int u^{-2} du = \frac{u^{-1}}{-1} = -\frac{1}{u} = -\frac{1}{x^2 + 1}$. Substituting these back,$I = 	an^{-1}(x) - (-\frac{1}{x^2 + 1}) + k = 	an^{-1}(x) + \frac{1}{x^2 + 1} + k$. Comparing this with the given form $	an^{-1}(x) + g(x) + k$,we find $g(x) = \frac{1}{x^2 + 1}$.

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

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