integral of 2x / (x^2 + 4x + 10) dx
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Hello, math!
Note the derivative of the denominator is: \(\displaystyle 2x\,+\,4\)
The fraction can be written: \(\displaystyle \L\frac{2x\,+\,4\,-\,4}{x^2\,+\,4x\,+\,10} \;=\;\frac{2x\,+\,4}{x^2\,+\,4x\,+\,10}\,-\,\frac{4}{x^2\,+\,4x\,+\,10}\)
Then we have: \(\displaystyle \L\:\int\frac{2x\,+\,4x}{x^2\,+\,4x\,+\,10}\,dx\:-\:4\int\frac{dx}{x^2\,+\,4x\,+\,10}\)
The first integral is: \(\displaystyle \L\,\int\frac{du}{u}\)
The second can be hammered into the \(\displaystyle arctan\) form:
. . \(\displaystyle x^2\,+\,4x\,+\,4\,+\,6 \:=\
x\,+\,2)^2\,+\,(\sqrt{6})^2\)
Note the derivative of the denominator is: \(\displaystyle 2x\,+\,4\)
The fraction can be written: \(\displaystyle \L\frac{2x\,+\,4\,-\,4}{x^2\,+\,4x\,+\,10} \;=\;\frac{2x\,+\,4}{x^2\,+\,4x\,+\,10}\,-\,\frac{4}{x^2\,+\,4x\,+\,10}\)
Then we have: \(\displaystyle \L\:\int\frac{2x\,+\,4x}{x^2\,+\,4x\,+\,10}\,dx\:-\:4\int\frac{dx}{x^2\,+\,4x\,+\,10}\)
The first integral is: \(\displaystyle \L\,\int\frac{du}{u}\)
The second can be hammered into the \(\displaystyle arctan\) form:
. . \(\displaystyle x^2\,+\,4x\,+\,4\,+\,6 \:=\
x\,+\,2)^2\,+\,(\sqrt{6})^2\)thanks, soroban!
i got the left side, but i'm still having trouble with the arctan part.
the derivative of arctan is 1/ (1 + x^2)
so the (x+2)^2 part makes sense to me because that is the x^2 part, but how is (sqrt(6)) the 1 part.
could you walk me through a little further for the tangent part?
thanks!
i got the left side, but i'm still having trouble with the arctan part.
the derivative of arctan is 1/ (1 + x^2)
so the (x+2)^2 part makes sense to me because that is the x^2 part, but how is (sqrt(6)) the 1 part.
could you walk me through a little further for the tangent part?
thanks!