Help proving integral reduction formula
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I'm having difficulties proving this reduction formula
If
$$I_{m,n}=intfrac{dx}{(ax+b)^m(cx+d)^n}$$
then
$$intfrac{dx}{(ax+b)^m(cx+d)^n}=-frac{1}{bc-ad}left[frac{1}{(ax+b)^{m-1}(cx+d)^{n-1}}+a(m+n-2)I_{m,n-1}right]$$
My attempt:
$$intfrac{dx}{(ax+b)^m(cx+d)^n}=intfrac{(cx+d)^{-n}}{(ax+b)^m}dx\
begin{vmatrix}u=frac{1}{(ax+b)^m}\
du=frac{-ma}{(ax+b)^{m+1}}dxend{vmatrix}v=int(cx+d)^{-n}dxquad v=frac{1}{c(1-n)(cx+d)^{n-1}}\
int udv=uv-int vdu\
=frac{1}{c(1-n)(ax+b)^m(cx+d)^{n-1}}+frac{ma}{c(1-n)}intfrac{dx}{(ax+b)^{m+1}(cx+d)^{n-1}}$$
This is where I'm stuck. Any help on this one?
calculus
asked Nov 15 at 20:52
Anson Pang
524
add a comment |
up vote
0
down vote
favorite
I'm having difficulties proving this reduction formula
If
$$I_{m,n}=intfrac{dx}{(ax+b)^m(cx+d)^n}$$
then
$$intfrac{dx}{(ax+b)^m(cx+d)^n}=-frac{1}{bc-ad}left[frac{1}{(ax+b)^{m-1}(cx+d)^{n-1}}+a(m+n-2)I_{m,n-1}right]$$
My attempt:
$$intfrac{dx}{(ax+b)^m(cx+d)^n}=intfrac{(cx+d)^{-n}}{(ax+b)^m}dx\
begin{vmatrix}u=frac{1}{(ax+b)^m}\
du=frac{-ma}{(ax+b)^{m+1}}dxend{vmatrix}v=int(cx+d)^{-n}dxquad v=frac{1}{c(1-n)(cx+d)^{n-1}}\
int udv=uv-int vdu\
=frac{1}{c(1-n)(ax+b)^m(cx+d)^{n-1}}+frac{ma}{c(1-n)}intfrac{dx}{(ax+b)^{m+1}(cx+d)^{n-1}}$$
This is where I'm stuck. Any help on this one?
calculus
asked Nov 15 at 20:52
Anson Pang
524
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm having difficulties proving this reduction formula
If
$$I_{m,n}=intfrac{dx}{(ax+b)^m(cx+d)^n}$$
then
$$intfrac{dx}{(ax+b)^m(cx+d)^n}=-frac{1}{bc-ad}left[frac{1}{(ax+b)^{m-1}(cx+d)^{n-1}}+a(m+n-2)I_{m,n-1}right]$$
My attempt:
$$intfrac{dx}{(ax+b)^m(cx+d)^n}=intfrac{(cx+d)^{-n}}{(ax+b)^m}dx\
begin{vmatrix}u=frac{1}{(ax+b)^m}\
du=frac{-ma}{(ax+b)^{m+1}}dxend{vmatrix}v=int(cx+d)^{-n}dxquad v=frac{1}{c(1-n)(cx+d)^{n-1}}\
int udv=uv-int vdu\
=frac{1}{c(1-n)(ax+b)^m(cx+d)^{n-1}}+frac{ma}{c(1-n)}intfrac{dx}{(ax+b)^{m+1}(cx+d)^{n-1}}$$
This is where I'm stuck. Any help on this one?
calculus
asked Nov 15 at 20:52
Anson Pang
524
I'm having difficulties proving this reduction formula
If
$$I_{m,n}=intfrac{dx}{(ax+b)^m(cx+d)^n}$$
then
$$intfrac{dx}{(ax+b)^m(cx+d)^n}=-frac{1}{bc-ad}left[frac{1}{(ax+b)^{m-1}(cx+d)^{n-1}}+a(m+n-2)I_{m,n-1}right]$$
My attempt:
$$intfrac{dx}{(ax+b)^m(cx+d)^n}=intfrac{(cx+d)^{-n}}{(ax+b)^m}dx\
begin{vmatrix}u=frac{1}{(ax+b)^m}\
du=frac{-ma}{(ax+b)^{m+1}}dxend{vmatrix}v=int(cx+d)^{-n}dxquad v=frac{1}{c(1-n)(cx+d)^{n-1}}\
int udv=uv-int vdu\
=frac{1}{c(1-n)(ax+b)^m(cx+d)^{n-1}}+frac{ma}{c(1-n)}intfrac{dx}{(ax+b)^{m+1}(cx+d)^{n-1}}$$
This is where I'm stuck. Any help on this one?
calculus
calculus
asked Nov 15 at 20:52
Anson Pang
524
asked Nov 15 at 20:52
Anson Pang
524
asked Nov 15 at 20:52
Anson Pang
524
asked Nov 15 at 20:52
Anson Pang
524
asked Nov 15 at 20:52
Anson Pang
524
524
add a comment |
add a comment |
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