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Find the integral: the fourth power of x plus 1, 1.
∫dx/( 1+x^4)

=∫dx/[(x^2+ax+ 1)(x^2-ax+ 1)]

=∫dx[( 1/2a)x+( 1/2)]/(x^2+ax+ 1)+∫dx[(- 1/2a)x+( 1/2)]/(x^2-ax+ 1)

=∫(x+a)dx/[2a(x^2+ax+ 1)]-∫(x-a)dx/[2a(x^2-ax+ 1)]

=∫(x+a/2+a/2)dx/[2a(x^2+ax+ 1)]-∫(x-a/2-a/2)dx/[2a(x^2-ax+ 1)]

=∫(x+a/2)dx/[2a(x^2+ax+ 1)]-∫(x-a/2)dx/[2a(x^2-ax+ 1)]+∫dx/[4(x^2+ax+ 1)]-∫adx/[4(x^2-ax+ 1)]

=( 1/2a)ln(x^2+ax+ 1)-( 1/2a)ln(x^2-ax+ 1)+( 1/2b)arctan(2x/b+a/b)-( 1/2b)arctan(2x/b-a/b)+c

A function is said to be integrable if its integral exists and is finite. Generally speaking, the integrand function does not necessarily have only one variable, and the integral domain can also be a space with different dimensions, even an abstract space without intuitive geometric significance. As mentioned above, for the real function f with only one variable x, the integral of f on the closed interval [a, b].

Extended data:

Under some definitions of integrals, these functions are not integrable, but under other definitions, their integrals exist. But sometimes due to teaching reasons, the definition will be different. The most common definitions of integral are Riemann integral and Lebesgue integral.

For a function f, if the riemann sum of the function f tends to a certain value S no matter how it is sampled and divided in the closed interval [a, b], as long as the maximum length of its subinterval is small enough, then the Riemann integral of f in the closed interval [a, b] exists and is defined as the limit S of riemann sum.

Integrals satisfy some basic properties. In the sense of Riemann integral, it represents an interval, and in the sense of Lebesgue integral, it represents a measurable set.

The integral of a function represents the overall properties of the function in a certain region, and changing the value of a certain point of the function will not change its integral value. For Riemannian integrable function, the integral remains unchanged by changing the value of the finite point. For Lebesgue integrable function, the change of function value on the set with measure 0 will not affect its integral value.

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