Integral of 1÷(1+x^2) dx
The solution
Detail solution
We have the integral:
/
|
| 1
| ------ dx
| 2
| 1 + x
|
/
Rewrite the integrand
1 1
------ = -------------
2 / 2 \
1 + x 1*\(-x) + 1/
or
/
|
| 1
| ------ dx
| 2 =
| 1 + x
|
/
/
|
| 1
| --------- dx
| 2
| (-x) + 1
|
/
In the integral
/
|
| 1
| --------- dx
| 2
| (-x) + 1
|
/
do replacement
then
the integral =
/
|
| 1
| ------ dv = atan(v)
| 2
| 1 + v
|
/
do backward replacement
/
|
| 1
| --------- dx = atan(x)
| 2
| (-x) + 1
|
/
Solution is:
The answer (Indefinite)
[src]
/
|
| 1
| ------ dx = C + atan(x)
| 2
| 1 + x
|
/
∫x2+11dx=C+atan(x)
The graph
Use the examples entering the upper and lower limits of integration.