Question

Verify the following indefinite integrals by differentiation. These integrals are derived in later chapters. $$\int \frac{x}{{(x^2-1)}^2}dx=-\frac{1}{2(x^2-1)}+C$$

Short answer

Question: Verify the correctness of the given indefinite integral by differentiation: $$\int \frac{x}{{(x^2-1)}^2}dx=-\frac{1}{2(x^2-1)}+C$$ Answer: The indefinite integral is verified to be correct, as the derivative of the antiderivative is equal to the original integrand: $$F^{\prime }(x)=\frac{x}{(x^2-1{)}^2}$$

Step-by-step solution

Write down the given integral and its antiderivative

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

Find the derivative of the antiderivative

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: $$F^{\prime }(x)=-\frac{1}{2}\cdot \frac{d}{dx}\left(\frac{1}{x^2-1}\right)$$ Now, we use the quotient rule for derivatives which is given by: $$\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v{u}^{\prime }-u{v}^{\prime }}{v^2}$$ Here, we set u=1 and v=x^2-1. Then, we can find the derivatives u' and v' as follows: $$u^{\prime }=\frac{d}{dx}(1)=0$$ $$v^{\prime }=\frac{d}{dx}(x^2-1)=2x$$ Apply the quotient rule for derivatives: $$F^{\prime }(x)=-\frac{1}{2}\cdot \frac{(-2x)(1)-(0)(x^2-1)}{(x^2-1{)}^2}$$ Now, simplify F'(x): $$F^{\prime }(x)=\frac{x}{(x^2-1{)}^2}$$

Compare the derivative with the original integrand

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