Intergration help (1 Viewer)

[c0okies]

hi thanks in advance for any help

View attachment 12165

using the substitution u = x - 1

View attachment 12166

using the substitution u = y + 1

[when a question doesnt contain the derivative but the sole pronumeral (as above) then what am i supposed to do?! ]

 

[.ben]

for the first one since u=x-1 therefore x=u+1

for the second one since u=y+1 therefore y=u-1

 

[pLuvia]

1

∫ [x/(x-3)³] dx

Let u = x-3 [x=u+3]
du=dx

.:∫ (u+3)/u³ du
= ∫ u^-2+3u^-3 du
= -1/u -3/2 u^-2 du
Sub u=x-3 back in
= -1/(x-3) - 3/2(x-3)^-2 + C

2

∫ y√y+1 dy

Let u=y+1 [y=u-1]
du=dy

.:∫ (u-1)u^1/2 du
= ∫ u^3/2-u^1/2 du
= 2/5 u^5/2 - 2/3 u^3/2 + C
Sub u=y+1 back in
= 2/5 (y+1)^5/2 - 2/3 (y+1)^3/2 + C

Hope that helped

 

[Riviet]

For the first one, instead of a substition of u=x-1, use u=x-3 and x=u+3
So integral of x/(x-3³)=(u+3)/u³du
=integral of u^-2du + integral of 3u^-3du and you can easily finish it off from there.

For the second, use u=y+1 and y=u-1
So integral of y.sqrt(y+1) = integral of (u-1).sqrtu du
= integral of u^3/2-u^-1/2 du, which you also finish off by youself.

Hope that helps.

 

[c0okies]

oic.. thanks alot; it helped quite alot =)