Question

find the indefinite integral:
∫(x)/(1+e^(-x^(2)))dx

I know this question can be solved by "integration by table", but ...can we solve it with General power rule? thanks

Answer 1

\[\int\limits \frac{ x }{ 1+e ^{-x ^{2}} }\]

Answer 2

simplify the expression first .
(remove negative exponent)

Answer 3

u sub?
try
u=x^2
-du/2=xdx
\[-\frac{1}{2} \int\limits \frac{du}{1+e^u} \]

Answer 4

well, if you simply first, then you'll find a better term to substitute for, try it.

Answer 5

simplify* first

Answer 6

\[ \frac{1}{1+e^u} = 1 - \frac{e^u}{1+e^u}\]
Also assuming u>0
\[ \frac{1}{1+e^u} = e^{-u}\sum_{n=0}^\infty (-1)^{n}e^{-nu}\]
if you integrate it .. you should get log series

Answer 7

my method is ...
first
\[(\frac{ x }{ 1+e ^{-x ^{2}} })(\frac{ e ^{x ^{2}} }{ e ^{x ^{2}} })\]
and then become...
\[\int\frac{ xe ^{x ^{2}} }{ e ^{x ^{2}}+1 }dx\]

\[{ u=e ^{x ^{2}}+1 }\]

...
but the ans looks not correct...
did I do something wrong?

Answer 8

tabluar integration
x+1|e^(x^2+2x)
1 |e^(x^2+2x)/(2x+2)
0 |e^(x^2+2x)/(2x+2)^2
y=(x+1)(e^(x^2+2x)/(2x+2))-(e^(x^2+2x)…

Answer 9

Let I =\[\int\limits \frac{ x }{ 1+e ^{-x ^{2}} }dx\]
\[substitute x ^{2}=t,diff. 2xdx=dt,xdx=\frac{ dt }{2 }\]
\[I=\frac{ 1 }{ 2 }\int\limits \frac{ dt }{1+e ^{-t} }\]
multiply numerator & denominator by e^t
\[I=\frac{ 1 }{ 2 }\int\limits \frac{ e ^{t} }{e ^{t}+1 }dt ,I=\frac{ 1 }{ 2 } \log \left( e ^{t}+1 \right)+c\]
substitute the value of t & get the answer.