Question

Evaluate the following integrals : $$\int \frac{x^{2}-1}{x^{2}+1} \cdot \frac{d x}{\sqrt{x^{4}+1}}$$

Short answer

Question: Evaluate the integral: $$\int \frac{x^{2}-1}{x^{2}+1} \cdot \frac{d x}{\sqrt{x^{4}+1}}$$ Answer: The value of the integral is: $$\frac{1}{4} \ln\left|\frac{x^4 + 2x^2 +2}{x^4 -x^2 +2}\right| + C$$

Step-by-step solution

Simplify the expression

Given expression: $$\int \frac{x^{2}-1}{x^{2}+1} \cdot \frac{d x}{\sqrt{x^{4}+1}}$$ Now, let's rewrite the expression in terms of the integrand: $$\int \frac{(x^{2} -1)(dx)}{(x^{2} +1)\sqrt{x^{4}+1}}$$

Choose a substitution method

We will use the substitution method to evaluate the integral, but we need to look for a suitable substitution to simplify the expression. Let \(u = x^2\). Then, \(du = 2xdx\) and \(dx = \frac{du}{2x}\). Keeping this in mind, let's substitute \(u\) in the given integral: $$\int \frac{(u -1)\frac{du}{2x}}{(u +1)\sqrt{u^2+1}}$$ Now, we have an integral of a simpler function: $$\frac{1}{2} \int \frac{(u -1) du}{(u +1)\sqrt{u^2+1}}$$

Evaluate the integral

To evaluate the integral, let's perform another substitution: Let \(v = u^2 +1\), then \(dv = 2u du\) and \(du = \frac{dv}{2u}\). The integral becomes: $$\frac{1}{2} \int \frac{(u -1)\frac{dv}{2u}}{(u +1)\sqrt{v}}$$ Which simplifies to: $$\frac{1}{4} \int \frac{(u -1) dv}{(u +1)\sqrt{v}}$$ Now, we substitute back \(u\) in terms of \(x\): $$\frac{1}{4} \int \frac{(x^2 -1) dv}{(x^2 +1)\sqrt{v}}$$ By evaluating the integral (notice that it's a standard form), we get: $$\frac{1}{4} \ln\left|\frac{v + (x^2 +1)}{v - (x^2 -1)}\right| + C$$

Write the final answer

Now let's substitute back \(v\) in terms of \(x\): $$\frac{1}{4} \ln\left|\frac{x^4 + 2x^2 +2}{x^4 -x^2 +2}\right| + C$$ The final answer is: $$\frac{1}{4} \ln\left|\frac{x^4 + 2x^2 +2}{x^4 -x^2 +2}\right| + C$$