Evaluate the integral: I = integral (sqrt(1+x²) +
Evaluate the integral: $$ I = \int \frac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1-x^4}} \, dx. $$
Explanation of Method
When a square root of a difference of fourth powers appears in a denominator, it often factors into a product of two simpler square roots via $a^2 - b^2 = (a-b)(a+b)$. Splitting the resulting fraction into separate terms then reduces the integral to standard forms involving $\arcsin$ and the inverse hyperbolic sine (logarithmic) integral.
- Factor the expression under the square root in the denominator.
- Split the fraction into two simpler fractions.
- Recognize each piece as a standard integral form.
- Combine the antiderivatives and add the constant of integration.
Step 1 · Simplify the square roots
The denominator is $\sqrt{1-x^4}$. Notice that:
$$ 1 - x^4 = (1-x^2)(1+x^2). $$
Thus, we can rewrite the denominator as:
$$ \sqrt{1-x^4} = \sqrt{(1-x^2)(1+x^2)} = \sqrt{1-x^2}\cdot\sqrt{1+x^2}. $$
Using this result, the integrand becomes:
$$ \frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1-x^4}} = \frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}\cdot\sqrt{1-x^2}}. $$
Separate the terms in the numerator:
$$ \frac{\sqrt{1+x^2}}{\sqrt{1+x^2}\cdot\sqrt{1-x^2}} + \frac{\sqrt{1-x^2}}{\sqrt{1+x^2}\cdot\sqrt{1-x^2}}. $$
Simplify each term:
$$ \frac{\sqrt{1+x^2}}{\sqrt{1+x^2}\cdot\sqrt{1-x^2}} = \frac{1}{\sqrt{1-x^2}}, \qquad \frac{\sqrt{1-x^2}}{\sqrt{1+x^2}\cdot\sqrt{1-x^2}} = \frac{1}{\sqrt{1+x^2}}. $$
Thus, the integral becomes:
$$ I = \int \left( \frac{1}{\sqrt{1-x^2}} + \frac{1}{\sqrt{1+x^2}} \right) dx. $$
Step 2 · Solve each term
The integral splits into two standard integrals:
$$ I = \int \frac{1}{\sqrt{1-x^2}}\,dx + \int \frac{1}{\sqrt{1+x^2}}\,dx. $$
First term: $\displaystyle \int \frac{1}{\sqrt{1-x^2}}\,dx$ is a standard integral:
$$ \int \frac{1}{\sqrt{1-x^2}}\,dx = \arcsin(x). $$
Second term: $\displaystyle \int \frac{1}{\sqrt{1+x^2}}\,dx$ is another standard integral:
$$ \int \frac{1}{\sqrt{1+x^2}}\,dx = \operatorname{arsinh}(x) = \ln\!\left(x+\sqrt{x^2+1}\right). $$
Step 3 · Combine the results
Combining both terms gives the final result:
$$ I = \arcsin(x) + \ln\!\left(x+\sqrt{x^2+1}\right) + C, $$
where $C$ is the constant of integration.
Summary Table
| Term | Standard form | Antiderivative |
|---|---|---|
| $1/\sqrt{1-x^2}$ | Standard $\arcsin$ integral | $\arcsin(x)$ |
| $1/\sqrt{1+x^2}$ | Standard $\operatorname{arsinh}$ integral | $\ln\!\left(x+\sqrt{x^2+1}\right)$ |
Additional Notes
- The domain requires $-1 \le x \le 1$ so that $1-x^4 \ge 0$ and both square roots in the original numerator are real.
- Treating $1-x^4$ as a difference of squares, $1^2-(x^2)^2=(1-x^2)(1+x^2)$, is the key factoring step that unlocks the simplification.
- The result can be checked by differentiating $\arcsin(x)+\ln\!\left(x+\sqrt{x^2+1}\right)$ and confirming it returns the original integrand.
Result
The solution to the integral is: