Question

∫2+3x^2/x^2(1+x^2)dx

Answer 1

-\frac{2}{x} + \arctan(x) + C

Answer 2

To solve the integral $\int \frac{2+3x^2}{x^2(1+x^2)}dx$, we use the method of partial fraction decomposition.

First, we decompose the integrand: $\frac{2+3x^2}{x^2(1+x^2)}$ Let $y = x^2$. Then the expression becomes: $\frac{2+3y}{y(1+y)}$ We can decompose this into partial fractions as: $\frac{2+3y}{y(1+y)} = \frac{A}{y} + \frac{B}{1+y}$ To find $A$ and $B$, we multiply both sides by $y(1+y)$: $2+3y = A(1+y) + By$ Set $y=0$: $2+3(0) = A(1+0) + B(0)$ $2 = A$ Set $y=-1$: $2+3(-1) = A(1-1) + B(-1)$ $2-3 = -B$ $-1 = -B$ $B = 1$ So, the decomposition in terms of $y$ is: $\frac{2+3y}{y(1+y)} = \frac{2}{y} + \frac{1}{1+y}$ Now, substitute back $y=x^2$: $\frac{2+3x^2}{x^2(1+x^2)} = \frac{2}{x^2} + \frac{1}{1+x^2}$ Now we can integrate term by term: $\int \frac{2+3x^2}{x^2(1+x^2)}dx = \int \left( \frac{2}{x^2} + \frac{1}{1+x^2} \right) dx$ $= \int 2x^{-2} dx + \int \frac{1}{1+x^2} dx$ Using the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$) and the standard integral $\int \frac{1}{1+x^2} dx = \arctan(x) + C$: $= 2 \left( \frac{x^{-2+1}}{-2+1} \right) + \arctan(x) + C$ $= 2 \left( \frac{x^{-1}}{-1} \right) + \arctan(x) + C$ $= -\frac{2}{x} + \arctan(x) + C$

Explanation of the solution: The integral of a rational function is solved using partial fraction decomposition. The integrand $\frac{2+3x^2}{x^2(1+x^2)}$ is simplified by substituting $y=x^2$ to $\frac{2+3y}{y(1+y)}$. This expression is then decomposed into $\frac{2}{y} + \frac{1}{1+y}$. Substituting $x^2$ back for $y$, we get $\frac{2}{x^2} + \frac{1}{1+x^2}$. Each term is then integrated separately using standard integration formulas: $\int x^{-2} dx = -x^{-1}$ and $\int \frac{1}{1+x^2} dx = \arctan(x)$.