Question

int[p(1+p)^(1/2)]

Answer 1

try \(u=1+p\)

Answer 2

\[\int\limits p(1+p)^{1/2} dp\]

Answer 3

u=1+p . du/dp = 1

Answer 4

i'm suppose to use integration by parts but I 'm getting a funny answer

Answer 5

parts? why not
\[\int p(\sqrt{1+p})dp\]
\[u=1+p, du=dp , p=u-1\]
\[\int (u-1)\sqrt{u}du\]
\[\int (u^{\frac{3}{2}}-u^{\frac{1}{2}})du\] etc

Answer 6

i know, i want to use integration by parts not substitution

Answer 7

ok so we want \(\int udv = uv-\int vdu\) what do we pick?

Answer 8

not really sure, boss

Answer 9

see i think it will come down to a u - sub using this method as well

Answer 10

we could say
\[u=p,dv = \sqrt{1+p}\] and then
\[du=dp\] and \[v=\int\sqrt{1+p}dp\]

Answer 11

yep

Answer 12

i guess that will work because
\[\int \sqrt{1+p}=2(1+p)^{\frac{3}{2}}\] of i am not mistaken

Answer 13

kk, go on..

Answer 14

so putting it together we get
\[uv-\int v du =2p(1+p)^{\frac{3}{2}}-\int 2(1+p)^{\frac{3}{2}}dp\] let me know if i messed up

Answer 15

\[2/3(1+p)^{3/2}\]

Answer 16

yeah right i forget the 3 in the denominator

Answer 17

\[uv-\int v du =\frac{2}{3}p(1+p)^{\frac{3}{2}}-\frac{2}{3}\int (1+p)^{\frac{3}{2}}dp\]should be better

Answer 18

yh

Answer 19

and the last one is ok by power rule, get
\[-\frac{3}{2}\int(1+p)^{\frac{3}{2}}dp=-\frac{15}{4}(1+p)^{\frac{5}{2}}\] unless i messed up again

Answer 20

no i messed up again

Answer 21

these \(p\)'s are getting on my nerves. what is wrong with \(x\)?

Answer 22

lol

Answer 23

\[-\frac{2}{3}\int(1+p)^{\frac{3}{2}}dp=-\frac{4}{15}(1+p)^{\frac{5}{2}}\]

Answer 24

there i think that is correct now

Answer 25

yh, prolly

Answer 26

first method is clearly snappier

Answer 27

kk, how do we wrap it up nicely by factorisation ? coz from what we did i got\[2/3[p(1+p)^{3/2}-2/5[(1+p)^{5/2}] +c\]

Answer 28

i will leave that up to you, because i have to bolt

Answer 29

hehe, no p. Thanks you, Master Shifu