r-soy
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Hi all
can please explaine to me what is integration of y/(x^2-y^2) dx
step by step ...
can please explaine to me what is integration of y/(x^2-y^2) dx
step by step ...
rsoy said:Hi all
can please explaine to me what is integration of y/(x^2-y^2) dx
step by step ...
rsoy said:1/2ln(x-y) + -1/2(x+y)
MarkFL said:Did you check out the Heaviside cover-up method I pointed you to the other day for partial fraction decomposition?
rsoy said:1/2ln(x-y) + -1/2ln(x+y) + c
rsoy said:Hi all
can please explaine to me what is integration of y/(x^2-y^2) dx
step by step ...
rsoy said:Hi all
can please explaine to me what is integration of y/(x^2-y^2) dx
step by step ...
I like Serena said:Another approach.
If you have a list of derivatives of trigonometric functions, you should have:
$$\frac{d}{dx} \text{ artanh } x = \frac 1 {1-x^2}$$
Since $y$ is treated as a constant $b$, we have: .$\displaystyle b\int\frac{dx}{x^2-b^2}$There is a standard integration formula: .$\displaystyle \int \frac{du}{u^2-a^2} \:=\:\frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right|+C $Therefore: .$\displaystyle b\left(\frac{1}{2b}\right)\ln\left|\frac{x-b}{x+b}\right|+C \;=\;\frac{1}{2}\ln\left|\frac{x-y}{x+y}\right|+C$$\displaystyle\int \frac{y}{x^2-y^2}\,dx$
soroban said:Hello, rsoy!
Since $y$ is treated as a constant $b$
Prove It said:The OP has not specified if this is actually the case...