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-
How to use it?
- Integral of d{x}:
-
Integral of (-1)/x^2
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Integral of 1/(x+2)^2
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Integral of x⁴
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Integral of x/(1+x²)
- Derivative of:
-
x/(1+x²)
-
Identical expressions
- x/(one +x²)
- x divide by (1 plus x²)
- x divide by (one plus x²)
- x/1+x²
- x divide by (1+x²)
- x/(1+x²)dx
-
Similar expressions
- (x+arctan(x)+x²arctan(x))/(1+x²)
- x/(1-x²)
- tan²x/1+x²
- (arctg³x)/(1+x²)
- (1-x)/(1+x²)
Integral of x/(1+x²) dx
The solution
You have entered
[src]
1 / | | x | ------ dx | 2 | 1 + x | / 0
$$\int\limits_{0}^{1} \frac{x}{x^{2} + 1}\, dx$$
Integral(x/(1 + x^2), (x, 0, 1))
Detail solution
We have the integral:
/ | | x | ------ dx | 2 | 1 + x | /
Rewrite the integrand
/ 2*x \
|------------| /0\
| 2 | |-|
x \x + 0*x + 1/ \1/
------ = -------------- + ---------
2 2 2
1 + x (-x) + 1or
/ | | x | ------ dx | 2 = | 1 + x | /
/
|
| 2*x
| ------------ dx
| 2
| x + 0*x + 1
|
/
------------------
2 In the integral
/
|
| 2*x
| ------------ dx
| 2
| x + 0*x + 1
|
/
------------------
2 do replacement
2 u = x
then
the integral =
/
|
| 1
| ----- du
| 1 + u
|
/ log(1 + u)
----------- = ----------
2 2 do backward replacement
/
|
| 2*x
| ------------ dx
| 2
| x + 0*x + 1
| / 2\
/ log\1 + x /
------------------ = -----------
2 2 In the integral
0
do replacement
v = -x
then
the integral =
True
do backward replacement
True
Solution is:
/ 2\
log\1 + x /
C + -----------
2
The answer (Indefinite)
[src]
/ | / 2\ | x log\1 + x / | ------ dx = C + ----------- | 2 2 | 1 + x | /
$$\int \frac{x}{x^{2} + 1}\, dx = C + \frac{\log{\left(x^{2} + 1 \right)}}{2}$$
The graph
The answer
[src]
log(2) ------ 2
$$\frac{\log{\left(2 \right)}}{2}$$
=
=
log(2) ------ 2
$$\frac{\log{\left(2 \right)}}{2}$$
log(2)/2
The graph
Use the examples entering the upper and lower limits of integration.