- #1
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Integral with sq. root in it...again
Find..
[tex]\int x^\frac{3}{2}\sqrt{1+x} dx[/tex]
Well I used the fact that:
[tex]\sqrt{1+x}=\sum_{n=0} ^\infty \frac{(-1)^n(2n!)x^n}{(1-2n)(n!)^24^n}[/tex]
and well I just multiplied by [itex]x^\frac{3}{2}[/itex]
so I integrated:
[tex]\int \sum_{n=0} ^\infty \frac{(-1)^n(2n!)x^(n+\frac{3}{2}}{(1-2n)(n!)^24^n}[/tex]
and got [tex]\sum_{n=0} ^\infty \frac{(-1)^n(2n!)x^(n+\frac{5}{2}}{(1-2n)(n!)^24^n\frac{5}{2}}[/tex]
[itex]\frac{2x^\frac{5}{2}}{5}\sqrt{1+x}[/itex] which is wrong because if i differentiate it I get an extra term in it
Homework Statement
Find..
[tex]\int x^\frac{3}{2}\sqrt{1+x} dx[/tex]
Homework Equations
The Attempt at a Solution
Well I used the fact that:
[tex]\sqrt{1+x}=\sum_{n=0} ^\infty \frac{(-1)^n(2n!)x^n}{(1-2n)(n!)^24^n}[/tex]
and well I just multiplied by [itex]x^\frac{3}{2}[/itex]
so I integrated:
[tex]\int \sum_{n=0} ^\infty \frac{(-1)^n(2n!)x^(n+\frac{3}{2}}{(1-2n)(n!)^24^n}[/tex]
and got [tex]\sum_{n=0} ^\infty \frac{(-1)^n(2n!)x^(n+\frac{5}{2}}{(1-2n)(n!)^24^n\frac{5}{2}}[/tex]
[itex]\frac{2x^\frac{5}{2}}{5}\sqrt{1+x}[/itex] which is wrong because if i differentiate it I get an extra term in it