Answer:
[tex]\int \left( \frac{X}{t-a}+\frac{Y}{t-b} \right)dt=\frac{1}{(b-a)} \left( -a\cdot log(t-a)+b\cdot log(t-a)\right) + C[/tex]
Step-by-step explanation:
[tex]\int \frac{tdt}{(t-a)(t-b)}[/tex] (1)
We can use partial fraction to solve this kind of integral.
In partial fraction we need to split the this fraction in two new fractions:
[tex]\frac{t}{(t-a)(t-b)}=\frac{X}{t-a}+\frac{Y}{t-b}[/tex] (2)
Now, we need to find the values of X and Y.
Let's work the right equation of (2). We can use least common denominator
[tex]\frac{X}{t-a}+\frac{Y}{t-b}=\frac{X(t-b)+Y(t-a)}{(t-a)(t-b)}=\frac{Xt-Xb+Yt-Ya}{(t-a)(t-b)}[/tex]
Now we can use common factor in the numerator.
[tex]\frac{X}{t-a}+\frac{Y}{t-b}=\frac{t(X+Y)-(Xb+Xa)}{(t-a)(t-b)}[/tex] (3)
If we see, we can compere the (3) whit the left side of (2).
The numerators are the same, so we can compare term by term, like this:
[tex]t=t(X+Y)[/tex] (4)
[tex]0=Xb+Xa[/tex] (5)
We just need to solve the system of equations ((4) and (5)) to find X and Y.
[tex]X=\frac{-a}{b-a}[/tex] and [tex]Y=\frac{b}{b-a}[/tex]
Now, putting this two values in (2) and next we will take the integral.
[tex]\int \left(\frac{X}{t-a}+\frac{Y}{t-b} \right)=\int \left(\frac{-a}{(b-a)(t-a)}+\frac{b}{(b-a)(t-b)} \right)=\frac{1}{(b-a)} \int \left(\frac{-a}{t-a}+\frac{b}{t-a}\right)dt[/tex]
Let's recall that the integral of (1/(x-a))dx is log(x-a), so using it we can find the integral.
[tex]\int \left(\frac{X}{t-a}+\frac{Y}{t-b} \right)dt=\frac{1}{(b-a)}\left( -a\cdot log(t-a)+b\cdot log(t-a)\right) + C[/tex]
I hope it helps you!
