Question

Evaluate the indefinite integral
\[\int\limits\frac{ 2y+3 }{ (y+7)^{3} }dy\]
I used u substitution u = y+7 and got
\[\frac{ -2(y-4) }{ (y+7)^{2} } + C\] would that be correct? :3

Answer 1

My work...
Evaluate the indefinite integral
\[\int\limits\frac{ 2y+3 }{ (y+7)^{3} }dy\]
u = y+7 -> y = u-7
du = 1
∫(2(u-7)+3)(u^-3)du
∫(2u-11)(u^-3)du
∫(2u^-2 - 11u^-3) du
-2u^-1 + 22u^-2 + C
-2u^-2 (u-11) + C
(-2 / (y+7)^2) (y+7-11) + C
\[\frac{ -2(y-4) }{ (y+7)^{2} } + C\]

Answer 2

what have you done after
\((2(u-7)+ 3 ) u^{-3}\)
?

Answer 3

distribute the 2
add like terms
distribute the u^-3

then take the integral ?

Answer 4

well, you didn't do it correctly

Answer 5

2(u-7)+3 = 2u -14 +3

Answer 6

isn't it ?

Answer 7

yeah
so 2u - 11

Answer 8

then distribute the u^-3 ?

Answer 9

oh, yeah that step is correct.

Answer 10

\[\int\limits \frac{ 2y+3 }{ \left( y+7 \right)^3 }dy=\int\limits \frac{ 2y+14-11 }{\left( y+7 \right)^3 }dy\]
\[=2\int\limits \left( y+7 \right)^{-2}dy-11\int\limits \left( y-7 \right)^{-3}dy\]
=?

Answer 11

ummm... so what'd I do wrong?

Answer 12

i don't think you went wrong anywhere, rechecking the resubstitution part

Answer 13

integral of 11u^(-3) is ?

Answer 14

11 u^(-2)/ (-2)
right ??

Answer 15

finally found the error after going it over 2-3 times :P

Answer 16

ohhhh okay, thanks :)

Answer 17

did you get it too ?
whats your new answer ?

Answer 18

\[\frac{ -2 }{ y+7 } + \frac{ 11 }{ 2(y+7)^{2} } + C\]?

Answer 19

-11(-1/2) u^-2
to 11 u^-2 / 2 ?

Answer 20

-4 (y+7) + 11 = -4y -28 +11 = -4y -17

you're correct :)

Answer 21

\(\frac{ -2 }{ y+7 } + \frac{ 11 }{ 2(y+7)^{2} } + C \quad \huge \checkmark\)

Answer 22

okay, THANK YOU!!! :D <3

Answer 23

welcome ^_^

Answer 24

correction i wrote y-7,it is y+7

Answer 25

yeah, got that ^_^ thanks for your help!

Answer 26

yw