Question

integral of x^3(1+x^2)^(3/2)dx

Answer 1

Hint. let u = x^2

Answer 2

no 1 +x^2=u

Answer 3

Hold on.

Answer 4

then seperate x^3 as x and x^2

Answer 5

\[(t-1)t ^{3/2}dt/2\]

Answer 6

\[\int\limits_{}^{}x^3(1+x^2)^\frac{ 3 }{ 2 }dx\]
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.

Answer 7

\[(t ^{5/2}-t ^{3/2})dt/2\]

Answer 8

answer=
\[t ^{7/2}/7-t ^{5/2}/5\]

Answer 9

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

Answer 10

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.

Answer 11

They all seem like different answers.

Answer 12

they are different ways to get the same answer

Answer 13

I reworked it and I ended up getting
\[\frac{ 2 }{ 14 }(1+x^2)^\frac{ 7 }{ 2 }- \frac{ 2 }{ 10 }(1+x^2)^\frac{ 5 }{ 2 }\]

Answer 14

I used only one substitution though.

Answer 15

That is right.