91Ó°ÊÓ

We have the given integral: $$\int \frac{x}{\left(x^{2}-1\right)^{2}} dx$$ And its antiderivative, with the integration constant C: $$F(x) = -\frac{1}{2\left(x^{2}-1\right)} + C$$

We need to find the derivative, F'(x), of the antiderivative function with respect to the variable x. For this, we'll use the chain rule and the quotient rule. The function can be rewritten as: $$F(x)= -\frac{1}{2} \cdot \frac{1}{(x^2-1)}+C$$ Let's differentiate F(x) with respect to x: $$F'(x)= -\frac{1}{2} \cdot \frac{d}{dx}\left(\frac{1}{x^2-1}\right)$$ Now, we use the quotient rule for derivatives which is given by: $$\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{vu'-uv'}{v^2}$$ Here, we set u=1 and v=x^2-1. Then, we can find the derivatives u' and v' as follows: $$u'= \frac{d}{dx} (1) = 0$$ $$v'= \frac{d}{dx} (x^2-1) = 2x$$ Apply the quotient rule for derivatives: $$F'(x)= -\frac{1}{2} \cdot \frac{(-2x)(1)-(0)(x^{2}-1)}{(x^{2}-1)^{2}}$$ Now, simplify F'(x): $$F'(x) = \frac{x}{(x^{2}-1)^{2}}$$

The derivative of the antiderivative, F'(x), is equal to the original integrand: $$F'(x) = \frac{x}{(x^{2}-1)^{2}}$$ Since both match, we can verify that the given indefinite integral is correct. The antiderivative of the integral indeed is: $$\int \frac{x}{\left(x^{2}-1\right)^{2}} dx = -\frac{1}{2\left(x^{2}-1\right)} + C$$