- #1
NINHARDCOREFAN
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The integral of [(x+2/(x+4)]^2
A/(x^2+4) + B/[(x^2+4)^2)
A=0, B=1
so, the integral of 1/(x^2+4)^2
how do you do this?
A/(x^2+4) + B/[(x^2+4)^2)
A=0, B=1
so, the integral of 1/(x^2+4)^2
how do you do this?
I think even when you don't do as what you said in your post, surely you can still find out A LOT OF SOURCES to get this THING solved. Why never look for books around you ? I am sure you should have a lot, because maths computation is what you like, which your presence here already tells me, RIGHT ?JonF said:Let u = 2tan and don't use partial fractions…
Hehe, Now your turn!NINHARDCOREFAN said:The integral of [(x+2/(x+4)]^2
A/(x^2+4) + B/[(x^2+4)^2)
A=0, B=1
so, the integral of 1/(x^2+4)^2
how do you do this?
Partial fractions are a mathematical technique used to simplify complex algebraic expressions by breaking them down into smaller, simpler fractions. This is often done in order to make solving equations or integrals easier.
You will typically need to use partial fractions when dealing with rational functions, which are functions that can be written as a ratio of two polynomials. When these functions are difficult to integrate or solve, partial fractions can be used to break them down into simpler fractions that are easier to work with.
The first step in solving equations with partial fractions is to decompose the original expression into simpler fractions using partial fraction decomposition. Then, you can solve for the unknown variables in each fraction and combine them back together to get the final solution.
Yes, there are a few rules to keep in mind when solving partial fractions. The first is that the degree of the numerator of each fraction must be less than the degree of the denominator. Additionally, if the denominator of a fraction can be factored, each factor should have its own corresponding term in the partial fraction decomposition.
Yes, partial fractions are commonly used in calculus when integrating rational functions. By breaking down the rational function into simpler fractions, it becomes easier to integrate and find the antiderivative. This is especially useful when dealing with improper fractions or functions with complex denominators.