Question

The integral $\int_{1/2025}^{2025} \frac{1+x^2}{x^2+x^{2022}}dx$ equals

Answer 1

2025 - \frac{1}{2025}

Answer 2

To evaluate the integral $I = \int_{1/2025}^{2025} \frac{1+x^2}{x^2+x^{2022}}dx$, we can use a standard property of definite integrals.

Let the integral be $I$. The limits of integration are of the form $a$ and $1/a$, where $a = 1/2025$. The property states that for an integral $\int_a^{1/a} f(x) dx$, we can make the substitution $x = 1/t$. Then $dx = -1/t^2 dt$. When $x = a$, $t = 1/a$. When $x = 1/a$, $t = a$.

Applying this substitution to $I$: $I = \int_{1/2025}^{2025} \frac{1+x^2}{x^2+x^{2022}}dx$ Substitute $x = 1/t$: $I = \int_{2025}^{1/2025} \frac{1+(1/t)^2}{(1/t)^2+(1/t)^{2022}} \left(-\frac{1}{t^2}\right) dt$ $I = -\int_{2025}^{1/2025} \frac{\frac{t^2+1}{t^2}}{\frac{1}{t^2}+\frac{1}{t^{2022}}} \frac{1}{t^2} dt$ $I = \int_{1/2025}^{2025} \frac{\frac{t^2+1}{t^2}}{\frac{t^{2020}+1}{t^{2022}}} \frac{1}{t^2} dt$ (changing limits and sign) $I = \int_{1/2025}^{2025} \frac{t^2+1}{t^2} \cdot \frac{t^{2022}}{t^{2020}+1} \cdot \frac{1}{t^2} dt$ $I = \int_{1/2025}^{2025} \frac{(t^2+1)t^{2018}}{t^{2020}+1} dt$

Now, replace $t$ with $x$: $I = \int_{1/2025}^{2025} \frac{(x^2+1)x^{2018}}{x^{2020}+1} dx$ (Equation 2)

We have the original integral: $I = \int_{1/2025}^{2025} \frac{1+x^2}{x^2+x^{2022}}dx = \int_{1/2025}^{2025} \frac{1+x^2}{x^2(1+x^{2020})}dx$ (Equation 1)

Add Equation 1 and Equation 2: $2I = \int_{1/2025}^{2025} \left( \frac{1+x^2}{x^2(1+x^{2020})} + \frac{(x^2+1)x^{2018}}{1+x^{2020}} \right) dx$ Factor out $(1+x^2)$ from the integrand: $2I = \int_{1/2025}^{2025} (1+x^2) \left( \frac{1}{x^2(1+x^{2020})} + \frac{x^{2018}}{1+x^{2020}} \right) dx$ Combine the fractions inside the parenthesis by finding a common denominator, which is $x^2(1+x^{2020})$: $\frac{1}{x^2(1+x^{2020})} + \frac{x^{2018} \cdot x^2}{x^2(1+x^{2020})} = \frac{1+x^{2020}}{x^2(1+x^{2020})}$ This simplifies to $\frac{1}{x^2}$.

Substitute this back into the integral: $2I = \int_{1/2025}^{2025} (1+x^2) \left( \frac{1}{x^2} \right) dx$ $2I = \int_{1/2025}^{2025} \left( \frac{1}{x^2} + \frac{x^2}{x^2} \right) dx$ $2I = \int_{1/2025}^{2025} \left( x^{-2} + 1 \right) dx$

Now, evaluate the integral: $2I = \left[ -\frac{1}{x} + x \right]_{1/2025}^{2025}$ $2I = \left( -\frac{1}{2025} + 2025 \right) - \left( -\frac{1}{1/2025} + \frac{1}{2025} \right)$ $2I = \left( -\frac{1}{2025} + 2025 \right) - \left( -2025 + \frac{1}{2025} \right)$ $2I = -\frac{1}{2025} + 2025 + 2025 - \frac{1}{2025}$ $2I = 2 \times 2025 - 2 \times \frac{1}{2025}$ $2I = 2 \left( 2025 - \frac{1}{2025} \right)$ Divide by 2 to find $I$: $I = 2025 - \frac{1}{2025}$