Question

Integration of 1/x²+x⁴

Answer 1

-(1/x) - arctan(x) + C

Answer 2

To integrate the given function, we first simplify the denominator and then use a suitable algebraic manipulation or partial fraction decomposition.

The given integral is: $I = \int \frac{1}{x^2+x^4} dx$

Step 1: Factor the denominator The denominator $x^2+x^4$ can be factored as $x^2(1+x^2)$. So the integral becomes: $I = \int \frac{1}{x^2(1+x^2)} dx$

Step 2: Decompose the integrand We can use a clever algebraic manipulation for this specific form. Notice that the numerator can be expressed in terms of the factors in the denominator: $1 = (1+x^2) - x^2$ Substitute this into the numerator of the integrand: $\frac{1}{x^2(1+x^2)} = \frac{(1+x^2) - x^2}{x^2(1+x^2)}$ Now, separate the fraction into two terms: $\frac{(1+x^2) - x^2}{x^2(1+x^2)} = \frac{1+x^2}{x^2(1+x^2)} - \frac{x^2}{x^2(1+x^2)}$ Simplify each term: $= \frac{1}{x^2} - \frac{1}{1+x^2}$

Step 3: Integrate term by term Now, integrate each term separately: $I = \int \left( \frac{1}{x^2} - \frac{1}{1+x^2} \right) dx$ $I = \int x^{-2} dx - \int \frac{1}{1+x^2} dx$ Recall the standard integration formulas: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$) $\int \frac{1}{1+x^2} dx = \arctan(x) + C$

Applying these formulas: $\int x^{-2} dx = \frac{x^{-2+1}}{-2+1} = \frac{x^{-1}}{-1} = -\frac{1}{x}$ $\int \frac{1}{1+x^2} dx = \arctan(x)$ Combining these results: $I = -\frac{1}{x} - \arctan(x) + C$ where $C$ is the constant of integration.

The final answer is $-\frac{1}{x} - \arctan(x) + C$.

Explanation: The problem involves integrating a rational function. The key steps are to factor the denominator and then use algebraic manipulation to split the fraction into simpler terms that can be integrated using standard formulas. In this case, recognizing that $1 = (1+x^2) - x^2$ allows for a quick decomposition without formal partial fractions. The integral of $x^{-2}$ is $-1/x$, and the integral of $1/(1+x^2)$ is $\arctan(x)$.