23-NOV

1. partial fraction decomposition
2. using 1. to compute integrals
3. integrals resulting in inverse trigonometric functions

1. Suppose we have a polynomial %20%3D%20x%5E2%2B4x%2B4%20%3D%20(x%2B2)%5E2#card=math&code=q%28x%29%20%3D%20x%5E2%2B4x%2B4%20%3D%20%28x%2B2%29%5E2&id=scxhm)

%20%3D%20%5Cfrac%7B4x-7%7D%7Bx%5E2%2B4x%2B4%7D%20%3D%20%20%5Cfrac%7BA%7D%7B(x%2B2)%5E2%7D%20%2B%20%5Cfrac%7BB%7D%7B(x%2B2)%7D#card=math&code=R%28x%29%20%3D%20%5Cfrac%7B4x-7%7D%7Bx%5E2%2B4x%2B4%7D%20%3D%20%20%5Cfrac%7BA%7D%7B%28x%2B2%29%5E2%7D%20%2B%20%5Cfrac%7BB%7D%7B%28x%2B2%29%7D&id=AFWjx) .

Clear the denominator: %20%3D%20Bx%2BA%2B2B#card=math&code=4x-7%20%3D%20A%20%2B%20B%28x%2B2%29%20%3D%20Bx%2BA%2B2B&id=Z3sUS) . So .

Now we can compute %20dx#card=math&code=%5Cint%20R%28x%29%20dx&id=NkFeB) :-).

Theorem of partial fraction decomposition:

For a rational function %3D%5Cfrac%7Bp(x)%7D%7Bq(x)%7D#card=math&code=R%28x%29%3D%5Cfrac%7Bp%28x%29%7D%7Bq%28x%29%7D&id=vaJyt) (for simplicity we can assume that %2C%20q(x)#card=math&code=p%28x%29%2C%20q%28x%29&id=Qig8U) are both monic, that is, the coefficients of the highest degree term in %2C%20q(x)#card=math&code=p%28x%29%2C%20q%28x%29&id=WbcJQ) respectively, are both ).

We assume that

%20%3D%20%5Cprod%7Bi%3D1%7D%5Em%20l_i(x)%5E%7B%5Cmu_i%7D%20%5Cprod%7Bj%3D1%7D%5En%20sj(x)%5E%7B%5Cnu_j%7D%0A#card=math&code=q%28x%29%20%3D%20%5Cprod%7Bi%3D1%7D%5Em%20li%28x%29%5E%7B%5Cmu_i%7D%20%5Cprod%7Bj%3D1%7D%5En%20s_j%28x%29%5E%7B%5Cnu_j%7D%0A&id=Uw69q)

is the factorisation of #card=math&code=q%28x%29&id=Qblj6) into irreducible polynomials, such that %3D1%2C%20%5Cdeg%20s_j(x)%3D2#card=math&code=%5Cdeg%20l_i%20%28x%29%3D1%2C%20%5Cdeg%20s_j%28x%29%3D2&id=GFbOi) with strictly negative discriminant. Then there is the following equation:

%3D%20%26%5Cfrac%7Bp(x)%7D%7Bq(x)%7D%5C%5C%0A%3D%20%26%5Cfrac%7BA%7B%5Cmu_1%7D%7D%7Bl_1(x)%5E%7B%5Cmu_1%7D%7D%20%2B%20%5Ccdots%20%2B%20%5Cfrac%7BA_1%7D%7Bl_1(x)%7D%20%2B%20%5Ccdots%2B%20%5Cfrac%7BB%7B%5Cmum%7D%7D%7Bl_m(x)%5E%7B%5Cmu_m%7D%7D%20%2B%20%5Cldots%20%2B%20%5Cfrac%7BB_1%7D%7Bl_m(x)%7D%20%5C%5C%0A%2B%26%5Cfrac%7BC%7B%5Cnu1%7Dx%2BD%7B%5Cnu1%7D%7D%7Bs_1(x)%5E%7B%5Cnu_1%7D%7D%2B%5Ccdots%20%2B%20%5Cfrac%7BC_1%20x%2BD_1%7D%7Bs_1(x)%7D%20%2B%20%5Ccdots%0A%5Cend%7Baligned%7D%0A#card=math&code=%5Cbegin%7Baligned%7D%0AR%28x%29%3D%20%26%5Cfrac%7Bp%28x%29%7D%7Bq%28x%29%7D%5C%5C%0A%3D%20%26%5Cfrac%7BA%7B%5Cmu1%7D%7D%7Bl_1%28x%29%5E%7B%5Cmu_1%7D%7D%20%2B%20%5Ccdots%20%2B%20%5Cfrac%7BA_1%7D%7Bl_1%28x%29%7D%20%2B%20%5Ccdots%2B%20%5Cfrac%7BB%7B%5Cmum%7D%7D%7Bl_m%28x%29%5E%7B%5Cmu_m%7D%7D%20%2B%20%5Cldots%20%2B%20%5Cfrac%7BB_1%7D%7Bl_m%28x%29%7D%20%5C%5C%0A%2B%26%5Cfrac%7BC%7B%5Cnu1%7Dx%2BD%7B%5Cnu_1%7D%7D%7Bs_1%28x%29%5E%7B%5Cnu_1%7D%7D%2B%5Ccdots%20%2B%20%5Cfrac%7BC_1%20x%2BD_1%7D%7Bs_1%28x%29%7D%20%2B%20%5Ccdots%0A%5Cend%7Baligned%7D%0A&id=glRFj)

1. Now we use this theorem, to discuss how to compute %20dx#card=math&code=%5Cint%20R%28x%29%20dx&id=MspiZ).

2.1 %3Dax%2Bb#card=math&code=l%28x%29%3Dax%2Bb&id=Ngzql) of multiplicity . (can assume that ).

%5E%5Cmu%7D#card=math&code=%5Cint%20%5Cfrac%7Bd%20x%7D%7B%28ax%2Bb%29%5E%5Cmu%7D&id=VSGPS) If , then %7D%7Bax%2Bb%7D%20%20%3D%20%5Cfrac%7B1%7D%7Ba%7D%20%5Cln%20%7Cax%2Bb%7C%20%2BC#card=math&code=%5Cint%20%5Cfrac%7Bd%20x%7D%7Bax%2Bb%7D%20%3D%20%5Cfrac%7B1%7D%7Ba%7D%20%5Cint%20%5Cfrac%7Bd%20%28ax%2Bb%29%7D%7Bax%2Bb%7D%20%20%3D%20%5Cfrac%7B1%7D%7Ba%7D%20%5Cln%20%7Cax%2Bb%7C%20%2BC&id=kGZ9W)

If , then %5E%5Cmu%7D%20%3D%20%5Cfrac%7B1%7D%7Ba%7D%20%5Cint%20(ax%2Bb)%5E%7B-%5Cmu%7Dd%20(ax%2Bb)%20%3D%20%5Cfrac%7B(ax%2Bb)%5E%7B1-%5Cmu%7D%7D%7Ba(1-%5Cmu)%7D%20%2BC#card=math&code=%5Cint%20%5Cfrac%7Bd%20x%7D%7B%28ax%2Bb%29%5E%5Cmu%7D%20%3D%20%5Cfrac%7B1%7D%7Ba%7D%20%5Cint%20%28ax%2Bb%29%5E%7B-%5Cmu%7Dd%20%28ax%2Bb%29%20%3D%20%5Cfrac%7B%28ax%2Bb%29%5E%7B1-%5Cmu%7D%7D%7Ba%281-%5Cmu%29%7D%20%2BC&id=YZI4h)

2.2 %3Dax%5E2%2Bbx%2Bc#card=math&code=s%28x%29%3Dax%5E2%2Bbx%2Bc&id=lP0L5) with of multiplicity .

We assume that #card=math&code=s%28x%29&id=Geeul) is monic. Then . %5E2%2Bc-b%5E2%2F4#card=math&code=x%5E2%2Bbx%2Bc%20%3D%20%28x%2Bb%2F2%29%5E2%2Bc-b%5E2%2F4&id=twbY8)

So we can turn to assume that the quadratic polynomial factors are all of the form and

%5E2%2Bt%7D%20%3D%20%5Cfrac%7BC%7D%7B2%7D%5Cfrac%7B(2x%2Bb)%2B2D-bC%2F2%7D%7B(x%2Bb%2F2)%5E2%2Bt%7D%20%3D%20%5Cfrac%7BC%7D%7B2%7D%20%5B%5Cfrac%7B2x%2Bb%7D%7B(x%2Bb%2F2)%5E2%7D%2B%20%5Cfrac%7B2D-bC%2F2%7D%7B(x%2Bb%2F2)%5E2%7D%5D#card=math&code=%5Cfrac%7BCx%2BD%7D%7B%28x%2Bb%2F2%29%5E2%2Bt%7D%20%3D%20%5Cfrac%7BC%7D%7B2%7D%5Cfrac%7B%282x%2Bb%29%2B2D-bC%2F2%7D%7B%28x%2Bb%2F2%29%5E2%2Bt%7D%20%3D%20%5Cfrac%7BC%7D%7B2%7D%20%5B%5Cfrac%7B2x%2Bb%7D%7B%28x%2Bb%2F2%29%5E2%7D%2B%20%5Cfrac%7B2D-bC%2F2%7D%7B%28x%2Bb%2F2%29%5E2%7D%5D&id=n3dNN)

If , then %7Ddx%20%3D%20%5Cint%20%5Cfrac%7Bd(x%5E2%2Bbx%2Bc)%7D%7B(x%5E2%2Bbx%2Bc)%7D%20%3D%20%5Cln%20%7Cx%5E2%2Bbx%2Bc%7C#card=math&code=%5Cint%20%5Cfrac%7B2x%2Bb%7D%7B%28x%5E2%2Bbx%2Bc%29%7Ddx%20%3D%20%5Cint%20%5Cfrac%7Bd%28x%5E2%2Bbx%2Bc%29%7D%7B%28x%5E2%2Bbx%2Bc%29%7D%20%3D%20%5Cln%20%7Cx%5E2%2Bbx%2Bc%7C&id=R9JwM) + Const.

The second term is %7D#card=math&code=%5Cint%20%5Cfrac%7Bdx%7D%7B%28x%5E2%2Bbx%2Bc%29%7D&id=aOovn). We alreday derived a formula for this integral last week.

If the multiplicity , then how to proceed?

%5E%7B%5Cnu%7D%7D%20%3D%20%5Cfrac%7BC%7D%7B2%7D%5Cfrac%7B(2x%2Bb)%2B2D-bC%2F2%7D%7B(x%5E2%2Bbx%2Bc)%5E%7B%5Cnu%7D%7D%20%3D%20%5Cfrac%7BC%7D%7B2%7D%20%5B%5Cfrac%7B2x%2Bb%7D%7B(x%5E2%2Bbx%2Bc)%5E%7B%5Cnu%7D%7D%2B%20%5Cfrac%7B2D-bC%2F2%7D%7B(x%5E2%2Bbx%2Bc)%5E%7B%5Cnu%7D%7D%5D#card=math&code=%5Cfrac%7BCx%2BD%7D%7B%28x%5E2%2Bbx%2Bc%29%5E%7B%5Cnu%7D%7D%20%3D%20%5Cfrac%7BC%7D%7B2%7D%5Cfrac%7B%282x%2Bb%29%2B2D-bC%2F2%7D%7B%28x%5E2%2Bbx%2Bc%29%5E%7B%5Cnu%7D%7D%20%3D%20%5Cfrac%7BC%7D%7B2%7D%20%5B%5Cfrac%7B2x%2Bb%7D%7B%28x%5E2%2Bbx%2Bc%29%5E%7B%5Cnu%7D%7D%2B%20%5Cfrac%7B2D-bC%2F2%7D%7B%28x%5E2%2Bbx%2Bc%29%5E%7B%5Cnu%7D%7D%5D&id=EYUi9)

%5E%5Cnu%7D%20dx%20%3D%20%5Cint%5Cfrac%7Bd(x%5E2%2Bbx%2Bc)%7D%7B(x%5E2%2Bbx%2Bc)%5E%7B%5Cnu%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B1-%5Cnu%7D%20(x%5E2%2Bbx%2Bc)%5E%7B1-%5Cnu%7D#card=math&code=%5Cint%20%5Cfrac%7B%202x%2Bb%7D%7B%28x%5E2%2Bbx%2Bc%29%5E%5Cnu%7D%20dx%20%3D%20%5Cint%5Cfrac%7Bd%28x%5E2%2Bbx%2Bc%29%7D%7B%28x%5E2%2Bbx%2Bc%29%5E%7B%5Cnu%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B1-%5Cnu%7D%20%28x%5E2%2Bbx%2Bc%29%5E%7B1-%5Cnu%7D&id=ieyba) + Const.

Using integration by parts: .

We take %5E%7B-%5Cnu%7D#card=math&code=v%20%3DX%2C%20u%3D%28X%5E2%2Bt%5E2%29%5E%7B-%5Cnu%7D&id=diEyB). Then %20%5Cfrac%7B2X%20d%20X%7D%7B(X%5E2%2Bt%5E2)%5E%7B%5Cnu%2B1%7D%7D#card=math&code=d%20u%20%3D%20%28-%5Cnu%29%20%5Cfrac%7B2X%20d%20X%7D%7B%28X%5E2%2Bt%5E2%29%5E%7B%5Cnu%2B1%7D%7D&id=riC4M) .

Then

So just need to know . In short, the recursive relation is

%5E%5Cnu%7D%20%2B%202%5Cnu%20I%5Cnu%20-%202%5Cnu%20t%5E2%20I%7B%5Cnu%2B1%7D%2C#card=math&code=Iv%20%3D%20%5Cfrac%7BX%7D%7B%28X%5E2%2Bt%5E2%29%5E%5Cnu%7D%20%2B%202%5Cnu%20I%5Cnu%20-%202%5Cnu%20t%5E2%20I_%7B%5Cnu%2B1%7D%2C&id=WGSGO) or equivalently

%5E%5Cnu%7D%20%2B%20%5Cfrac%7B2%5Cnu%20-%201%7D%7B2%5Cnu%20t%5E2%7DI%5Cnu.%0A#card=math&code=I%7B%5Cnu%2B1%7D%20%3D%20%5Cfrac%7B1%7D%7B2%5Cnu%20t%5E2%7D%20%5Cfrac%7BX%7D%7B%28X%5E2%2Bt%5E2%29%5E%5Cnu%7D%20%2B%20%5Cfrac%7B2%5Cnu%20-%201%7D%7B2%5Cnu%20t%5E2%7DI_%5Cnu.%0A&id=Rb37V)

Note that , which we already know.

1. Assume that . If one sees that , then firstly take out the constant :

Then we use the Pythagorean identity: .

Trigonometric (sine) substitution: .

%20%3D%20a%20%5Ccos%20%5Ctheta%20d%20%5Ctheta#card=math&code=dx%20%3D%20a%20d%28%5Csin%20%5Ctheta%29%20%3D%20a%20%5Ccos%20%5Ctheta%20d%20%5Ctheta&id=cWWsK)

%20%2BC#card=math&code=%5Cint%20%5Cfrac%7Bdx%7D%7B%5Csqrt%7Ba%5E2-x%5E2%7D%7D%20%3D%20%5Cint%20%5Cfrac%7Ba%20%5Ccos%20%5Ctheta%20d%20%5Ctheta%7D%7Ba%20%5Ccos%20%5Ctheta%7D%20%3D%20%5Cint%201%20d%20%5Ctheta%20%3D%20%5Ctheta%20%2BC%20%3D%20%5Carcsin%20%28x%2Fa%29%20%2BC&id=SHUNa)

Another trigonometric identity:

%5E2)#card=math&code=a%5E2%2Bx%5E2%20%3D%20a%5E2%281%2B%28x%2Fa%29%5E2%29&id=ngB2G) we let , then .

%5E2%7D%20%3D%20%5Cfrac%7B1%7D%7Ba%5E2%7D%20%5Cint%20%5Cfrac%7Ba%20d%20%5Ctheta%2F%5Ccos%5E2%20%20%5Ctheta%7D%7B1%2F%5Ccos%5E2%20%5Ctheta%7D%20%3D%20%5Cfrac%7B1%7D%7Ba%7D%20%5Cint%201%20d%20%5Ctheta%20%3D(1%2Fa)%20%5Ctheta%20%2B%20C%20%3D%20(1%2Fa)%5Carctan(x%2Fa)%2BC#card=math&code=%5Cint%20%5Cfrac%7Bdx%7D%7Ba%5E2%2Bu%5E2%7D%20%3D%20%5Cfrac%7B1%7D%7Ba%5E2%7D%20%5Cint%20%5Cfrac%7B%20d%20x%7D%7B1%2B%28x%2Fa%29%5E2%7D%20%3D%20%5Cfrac%7B1%7D%7Ba%5E2%7D%20%5Cint%20%5Cfrac%7Ba%20d%20%5Ctheta%2F%5Ccos%5E2%20%20%5Ctheta%7D%7B1%2F%5Ccos%5E2%20%5Ctheta%7D%20%3D%20%5Cfrac%7B1%7D%7Ba%7D%20%5Cint%201%20d%20%5Ctheta%20%3D%281%2Fa%29%20%5Ctheta%20%2B%20C%20%3D%20%281%2Fa%29%5Carctan%28x%2Fa%29%2BC&id=BswFS)