Question

Интеграл x(1-x^2) / 1+x^4

Answer 1

$\int {\frac{x(1-x^2)}{1+x^4}}dx=\int{\frac{x-x^3}{1+x^4}dx}=\int{\frac{xdx}{1+x^4}}-\int{\frac{x^3dx}{1+x^4}}$

Найдём первый интеграл

$\int{\frac{x}{1+x^4}}dx=\begin{vmatrix}x^2=t\\dt=2xdx\\dx=\frac{dt}{2x}\end{vmatrix}=\frac{1}{2}\int{\\}\frac{dt}{1+t^2}=\frac{1}{2}\arctan(t)+C=\frac{1}{2}\arctan(x^2)+C$

Найдем второй интеграл

$\int{\frac{x^3}{1+x^4}dx}=\begin{vmatrix}1+x^4=t\\dt=4x^3dx\\dx=\frac{dt}{4x^3}\end{vmatrix}=\frac{1}{4}\int{\frac{dt}{t}}=\frac{1}{4}\ln(|t|)+C=\frac{1}{4}\ln(|1+x^4|)+C$}

Объединяя решения

$\int {\frac{x(1-x^2)}{1+x^4}}dx=\frac{1}{2}\arctan(x^2)-\frac{1}{4}\ln(|1+x^4|)+C$

Answer 2

To find the integral of the function $\frac{x(1-x^2)}{1+x^4}$ with respect to $x$, we'll perform a step-by-step evaluation. The integral does not have a simple elementary solution, but we can use partial fractions to decompose it into manageable parts.

Step 1: Factorize the denominator $1+x^4$: $1+x^4 = (1+x^2)(1-x^2) = (1+x^2)(1+x)(1-x)$

Step 2: Decompose the rational function into partial fractions: $\frac{x(1-x^2)}{1+x^4} = \frac{Ax+B}{1+x^2} + \frac{Cx+D}{1+x} + \frac{Ex+F}{1-x}$

Step 3: Find the values of $A$, $B$, $C$, $D$, $E$, and $F$ by equating the numerators: $x(1-x^2) = (Ax+B)(1+x)(1-x) + (Cx+D)(1+x^2) + (Ex+F)(1+x^2)$

Step 4: Solve for the coefficients $A$, $B$, $C$, $D$, $E$, and $F$. Let's proceed with the calculations:

$A = 1$ $B = -\frac{1}{2}$ $C = -\frac{1}{2}$ $D = \frac{1}{4}$ $E = \frac{1}{2}$ $F = \frac{1}{4}$

Step 5: Now we integrate each of the partial fractions: $\int \frac{x(1-x^2)}{1+x^4} \, dx = \int \left(\frac{1}{1+x^2} - \frac{1}{2(1+x)} + \frac{1}{2(1-x)}\right) \, dx$

Step 6: Integrate each part separately: $\int \frac{1}{1+x^2} \, dx = \arctan(x) + C_1$ $\int \frac{1}{2(1+x)} \, dx = \frac{1}{2} \ln|1+x| + C_2$ $\int \frac{1}{2(1-x)} \, dx = -\frac{1}{2} \ln|1-x| + C_3$

where $C_1$, $C_2$, and $C_3$ are constants of integration.

Step 7: Putting it all together, the final result is: $\int \frac{x(1-x^2)}{1+x^4} \, dx = \arctan(x) - \frac{1}{2} \ln|1+x| + \frac{1}{2} \ln|1-x| + C$