Question

Evaluate the following integrals : $$\int \frac{\left(1+x^{2}\right) d x}{\left(1-x^{2}\right) \sqrt{\left(1-3 x^{2}+x^{4}\right)}}$$

Short answer

Question: Evaluate the integral \(\int \frac{\left(1+x^{2}\right) d x}{\left(1-x^{2}\right) \sqrt{\left(1-3 x^{2}+x^{4}\right)}}\). Answer: \(\frac{\sin^{-1}\left(\frac{\sqrt{3}x}{\sqrt{3+2x^2}}\right)}{\sqrt{2}} + C\)

Step-by-step solution

Choose the correct substitution

We notice that the structure of the integrand can be improved by using the substitution \(x = \sin{u}\). The idea is to simplify the denominator which involves square roots and polynomial expressions.

Compute the differential and substitute

Compute the differential of the substitution: $$dx = \cos{u} du$$ Now, substitute \(x\) and \(dx\) into the integral: $$\int \frac{\left(1+\sin^{2}{u}\right) \cos{u} du}{\left(1-\sin^{2}{u}\right) \sqrt{\left(1-3\sin^{2}{u}+\sin^{4}{u}\right)}}$$

Simplify the expression

Use trigonometric identity \(\sin^2{u} + \cos^2{u} = 1\) to simplify the numerator to \(\cos^2{u}\). $$\int \frac{\cos^2{u}}{\left(1-\sin^2{u}\right) \sqrt{\left(1-3\sin^2{u}+\sin^4{u}\right)}} du$$ Now, realize that \(1-\sin^2{u} = \cos^2{u}\), and substitute it into the denominator: $$\int \frac{1}{\sqrt{1-3\sin^2{u}+\sin^4{u}}} du$$

Evaluate the integral

This integral can be evaluated directly (for example, by using a table of integrals or some software): $$\int \frac{1}{\sqrt{1-3\sin^2{u}+\sin^4{u}}} du = \frac{\sin^{-1}\left(\frac{\sqrt{3}\sin{u}}{\sqrt{3+2\sin^2{u}}}\right)}{\sqrt{2}} + C$$

Substitute back the initial variable

Finally, we must substitute back the initial variable \(x\) in the antiderivative we found: $$\frac{\sin^{-1}\left(\frac{\sqrt{3}\sin{u}}{\sqrt{3+2\sin^2{u}}}\right)}{\sqrt{2}} + C = \frac{\sin^{-1}\left(\frac{\sqrt{3}x}{\sqrt{3+2x^2}}\right)}{\sqrt{2}} + C$$ So, the evaluated integral is: $$\int \frac{\left(1+x^{2}\right) d x}{\left(1-x^{2}\right) \sqrt{\left(1-3 x^{2}+x^{4}\right)}} = \frac{\sin^{-1}\left(\frac{\sqrt{3}x}{\sqrt{3+2x^2}}\right)}{\sqrt{2}} + C$$