Question

evaluate the indefinite integral......x^9/ sqrt(3+x^5)..... i got pretty far but now i need help because im stuck at a certain point

Answer 1

x^9
----------- this thing?
sqrt(3+x^5)

Answer 2

im stuck at (1/5) integrate (u-3) /sqrt(u)..................

Answer 3

x^9 * sqrt(3+x^5)
---- ----------
sqrt(3+x^5) sqrt(3+x^5)

=x^9

Answer 4

yes ha

Answer 5

x^9
___________
sqrt3 +x^5

Answer 6

i got to the point where i have a split integration

Answer 7

x^9
----------- u = 3+x^5; du = 5x^4 dx; du/5x^4 = dx
sqrt(3+x^5)



x^9 du x^5 du
----------- = ---------
5x^4 u^(1/2) 5 u^(1/2)

what x=?

u = 3+x^5; x^5 = u-3 right


u-3 du
---------
5 u^(1/2)

Answer 8

\[\int\limits_{} \frac{1}{5} u(u^{-1/2}) - \int\limits_{} \frac{3}{5} u^{-1/2}\]

Answer 9

thats where im at now .. then you have to split them and get the integral of u / 5 u ^(1/2) - 3/ 5 u ^(1/2)

Answer 10

u[u^(-1/2)] = u^(1/2) right?

Answer 11

\[\frac{1}{5} \int\limits_{} u^{1/2} - \frac{3}{5} \int\limits_{} u^{-1/2}\]

Answer 12

wouldnt it be (2/5)?

Answer 13

wouldnt it be (2/5)?

Answer 14

for the first one that is ../.

Answer 15

because to get rid of the half in the denominator you have to multiply by 2?

Answer 16

\[\frac{u^{(\frac{1}{2} + 1)}}{5(\frac{1}{2} + 1)} - \frac{3u^{(-\frac{1}{2}+1})}{5(- \frac{1}{2}+1)}+C\]

Answer 17

there is no 1/2 in the denominator; thats an exponent :)

Answer 18

ahhh pellet i see now ... so what now after thats ?

Answer 19

\[\frac{2u \sqrt{u}}{15} - \frac{6}{5\sqrt{u}} + C\] if I kept track of everything..its easier on paper i spose :)

Answer 20

soooo close :) move that sqrt(u) back up top with the "6"

Answer 21

change all yor "u"s back into (x+x^5)

Answer 22

how did you get 2U?

Answer 23

the bottom of that side goes to 5(3/2) = 15/2...flip the 2 on top..... shortcut math really; the proper what is to multiply bythe reciprocal of the denominator

Answer 24

Possible intermediate steps:
integral x^9/sqrt(3+x^5) dx
For the integrand x^9/sqrt(x^5+3), substitute u = x^5 and du = 5 x^4 dx:
= 1/5 integral u/sqrt(u+3) du
For the integrand u/sqrt(u+3), substitute s = u+3 and ds = du:
= 1/5 integral (s-3)/sqrt(s) ds
Expanding the integrand (s-3)/sqrt(s) gives sqrt(s)-3/sqrt(s):
= 1/5 integral (sqrt(s)-3/sqrt(s)) ds
Integrate the sum term by term and factor out constants:
= 1/5 integral sqrt(s) ds-3/5 integral 1/sqrt(s) ds
The integral of 1/sqrt(s) is 2 sqrt(s):
= 1/5 integral sqrt(s) ds-(6 sqrt(s))/5
The integral of sqrt(s) is (2 s^(3/2))/3:
= (2 s^(3/2))/15-(6 sqrt(s))/5+constant
Substitute back for s = u+3:
= 2/15 (u+3)^(3/2)-(6 sqrt(u+3))/5+constant
Substitute back for u = x^5:
= 2/15 (x^5+3)^(3/2)-(6 sqrt(x^5+3))/5+constant
Which is equal to:
= 2/15 (x^5-6) sqrt(x^5+3)+constant

Answer 25

and sqrt(u^3) = u sqrt(u)

Answer 26

wow thank you so much ...