Question

Calculate the indefinite integral: (Assume t> a and t > b. Use C for the constant of integration.)

Answer 1

Answer:

$\int \left( \frac{X}{t-a}+\frac{Y}{t-b} \right)dt=\frac{1}{(b-a)} \left( -a\cdot log(t-a)+b\cdot log(t-a)\right) + C$

Step-by-step explanation:  

$\int \frac{tdt}{(t-a)(t-b)}$ (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:

$\frac{t}{(t-a)(t-b)}=\frac{X}{t-a}+\frac{Y}{t-b}$ (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

$\frac{X}{t-a}+\frac{Y}{t-b}=\frac{X(t-b)+Y(t-a)}{(t-a)(t-b)}=\frac{Xt-Xb+Yt-Ya}{(t-a)(t-b)}$

Now we can use common factor in the numerator.

$\frac{X}{t-a}+\frac{Y}{t-b}=\frac{t(X+Y)-(Xb+Xa)}{(t-a)(t-b)}$ (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:

$t=t(X+Y)$ (4)

$0=Xb+Xa$ (5)

We just need to solve the system of equations ((4) and (5)) to find X and Y.

$X=\frac{-a}{b-a}$ and $Y=\frac{b}{b-a}$

Now, putting this two values in (2) and next we will take the integral.

$\int \left(\frac{X}{t-a}+\frac{Y}{t-b} \right)=\int \left(\frac{-a}{(b-a)(t-a)}+\frac{b}{(b-a)(t-b)} \right)=\frac{1}{(b-a)} \int \left(\frac{-a}{t-a}+\frac{b}{t-a}\right)dt$

Let's recall that the integral of (1/(x-a))dx is log(x-a), so using it we can find the integral.

$\int \left(\frac{X}{t-a}+\frac{Y}{t-b} \right)dt=\frac{1}{(b-a)}\left( -a\cdot log(t-a)+b\cdot log(t-a)\right) + C$

I hope it helps you!