anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
x^9 ----------- this thing? sqrt(3+x^5)
anonymous
  • anonymous
im stuck at (1/5) integrate (u-3) /sqrt(u)..................
anonymous
  • anonymous
x^9 * sqrt(3+x^5) ---- ---------- sqrt(3+x^5) sqrt(3+x^5) =x^9

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anonymous
  • anonymous
yes ha
anonymous
  • anonymous
x^9 ___________ sqrt3 +x^5
anonymous
  • anonymous
i got to the point where i have a split integration
amistre64
  • amistre64
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)
amistre64
  • amistre64
\[\int\limits_{} \frac{1}{5} u(u^{-1/2}) - \int\limits_{} \frac{3}{5} u^{-1/2}\]
anonymous
  • anonymous
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)
amistre64
  • amistre64
u[u^(-1/2)] = u^(1/2) right?
amistre64
  • amistre64
\[\frac{1}{5} \int\limits_{} u^{1/2} - \frac{3}{5} \int\limits_{} u^{-1/2}\]
anonymous
  • anonymous
wouldnt it be (2/5)?
anonymous
  • anonymous
wouldnt it be (2/5)?
anonymous
  • anonymous
for the first one that is ../.
anonymous
  • anonymous
because to get rid of the half in the denominator you have to multiply by 2?
amistre64
  • amistre64
\[\frac{u^{(\frac{1}{2} + 1)}}{5(\frac{1}{2} + 1)} - \frac{3u^{(-\frac{1}{2}+1})}{5(- \frac{1}{2}+1)}+C\]
amistre64
  • amistre64
there is no 1/2 in the denominator; thats an exponent :)
anonymous
  • anonymous
ahhh pellet i see now ... so what now after thats ?
amistre64
  • amistre64
\[\frac{2u \sqrt{u}}{15} - \frac{6}{5\sqrt{u}} + C\] if I kept track of everything..its easier on paper i spose :)
amistre64
  • amistre64
soooo close :) move that sqrt(u) back up top with the "6"
amistre64
  • amistre64
change all yor "u"s back into (x+x^5)
anonymous
  • anonymous
how did you get 2U?
amistre64
  • amistre64
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
anonymous
  • anonymous
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
amistre64
  • amistre64
and sqrt(u^3) = u sqrt(u)
anonymous
  • anonymous
wow thank you so much ...

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