anonymous
  • anonymous
integral of x^3(1+x^2)^(3/2)dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Hint. let u = x^2
anonymous
  • anonymous
no 1 +x^2=u
anonymous
  • anonymous
Hold on.

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anonymous
  • anonymous
then seperate x^3 as x and x^2
anonymous
  • anonymous
\[(t-1)t ^{3/2}dt/2\]
anonymous
  • anonymous
\[\int\limits_{}^{}x^3(1+x^2)^\frac{ 3 }{ 2 }dx\] u = x^2 du = 2xdx; dx = du/2x \[\frac{ 1 }{ 2 }\int\limits_{?}^{?}u(1+u)^\frac{ 3 }{ 2 }du\] substitute s = 1+u ds = du \[\frac{ 1 }{ 2 }\int\limits_{}^{}(-1+s)s^\frac{ 3 }{ 2}ds\] expand integrate and substitue from there.
anonymous
  • anonymous
\[(t ^{5/2}-t ^{3/2})dt/2\]
anonymous
  • anonymous
answer= \[t ^{7/2}/7-t ^{5/2}/5\]
anonymous
  • anonymous
Wolframalpha has something different: For the integrand x^3 (1+x^2)^(3/2), substitute u = x^2 and du = 2 x dx: = 1/2 integral u (1+u)^(3/2) du For the integrand u (1+u)^(3/2), substitute s = 1+u and ds = du: = 1/2 integral (-1+s) s^(3/2) ds Expanding the integrand (-1+s) s^(3/2) gives -s^(3/2)+s^(5/2): = 1/2 integral (-s^(3/2)+s^(5/2)) ds Integrate the sum term by term and factor out constants: = -1/2 integral s^(3/2) ds+1/2 integral s^(5/2) ds The integral of s^(3/2) is (2 s^(5/2))/5: = -s^(5/2)/5+1/2 integral s^(5/2) ds The integral of s^(5/2) is (2 s^(7/2))/7: = -s^(5/2)/5+s^(7/2)/7+constant Substitute back for s = 1+u: = -1/5 (1+u)^(5/2)+1/7 (1+u)^(7/2)+constant Substitute back for u = x^2: = -1/5 (1+x^2)^(5/2)+1/7 (1+x^2)^(7/2)+constant Which is equal to: Answer: | | = 1/35 (1+x^2)^(5/2) (-2+5 x^2)+constant
anonymous
  • anonymous
That's how I got mine :p Aditya had a better answer though because he only used substitution one time. You can use either method.
anonymous
  • anonymous
They all seem like different answers.
anonymous
  • anonymous
they are different ways to get the same answer
anonymous
  • anonymous
I reworked it and I ended up getting \[\frac{ 2 }{ 14 }(1+x^2)^\frac{ 7 }{ 2 }- \frac{ 2 }{ 10 }(1+x^2)^\frac{ 5 }{ 2 }\]
anonymous
  • anonymous
I used only one substitution though.
anonymous
  • anonymous
That is right.

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