Integration by Parts: Solving Integrals without Laplace Transforms

[shashankac655]

1. How to solve integral of (1/(t²-t))dt
2. to be solved without using laplace transforms
3. integral of( uv)= u*(integral of v) -integral of ((u')*(integral of v)) ... right?

integral of (1/t^2-t) = integral of (1/t)*(1/t-1)dt = (1/t-1)*(log t) - integral((-1/(t-1)²*logt ...i don't know how to continue.

 

[shakgoku]

hi shashankac655!

Most straightforward way to solve it is using partial fractions.

 

[Mark44]

Using shakgoku's suggestion, break up 1/(t² - t) into A/t + B/(t - 1) and solve for the constants A and B for which the equation 1/(t² - t) = A/t + B/(t - 1) is an identity (except for t = 0 and t = 1).

 

[SammyS]

1/(t2 - t) = A/t + B/(t - 1)

Multiply both sides by t(t - 1). The resulting equation is easy to solve for A & B by setting t = 0 to get A and then t = 1 to get B. This works for this equation since both sides are defined for all t, whereas the previous equation has expressions which are undefined for t = 0 and t = 1.

 

[shashankac655]

ok thanks...i don't know why i didn't think of that!

 

[Dickfore]

Or, you can complete the square:

<br /> t^{2} - t = t^{2} - 2 t \, \frac{1}{2} + \left(\frac{1}{2}\right)^{2} - \left(\frac{1}{2}\right)^{2} = \left(t - \frac{1}{2}\right)^{2}- \left(\frac{1}{2}\right)^{2}<br />

and use the hyperbolic substitution:

<br /> t - \frac{1}{2} = \frac{1}{2} \, \cosh{u}<br />