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1/(x²+1)²
Solving integrals/ 1/(x²+1)²

Integral of 1/(x²+1)² dx

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

You have entered [src] 
  1             
  /             
 |              
 |      1       
 |  --------- dx
 |          2   
 |  / 2    \    
 |  \x  + 1/    
 |              
/               
0
$$\int\limits_{0}^{1} \frac{1}{\left(x^{2} + 1\right)^{2}}\, dx$$ 
Integral(1/((x^2 + 1)^2), (x, 0, 1))
Detail solution

    TrigSubstitutionRule(theta=_theta, func=tan(_theta), rewritten=cos(_theta)**2, substep=RewriteRule(rewritten=cos(2*_theta)/2 + 1/2, substep=AddRule(substeps=[ConstantTimesRule(constant=1/2, other=cos(2*_theta), substep=URule(u_var=_u, u_func=2*_theta, constant=1/2, substep=ConstantTimesRule(constant=1/2, other=cos(_u), substep=TrigRule(func='cos', arg=_u, context=cos(_u), symbol=_u), context=cos(_u), symbol=_u), context=cos(2*_theta), symbol=_theta), context=cos(2*_theta)/2, symbol=_theta), ConstantRule(constant=1/2, context=1/2, symbol=_theta)], context=cos(2*_theta)/2 + 1/2, symbol=_theta), context=cos(_theta)**2, symbol=_theta), restriction=True, context=1/((x**2 + 1)**2), symbol=x)

  1. Now simplify:

  2. Add the constant of integration:

The answer is:

The answer (Indefinite) [src] 
  /                                       
 |                                        
 |     1              atan(x)       x     
 | --------- dx = C + ------- + ----------
 |         2             2        /     2\
 | / 2    \                     2*\1 + x /
 | \x  + 1/                               
 |                                        
/
$$\int \frac{1}{\left(x^{2} + 1\right)^{2}}\, dx = C + \frac{x}{2 \left(x^{2} + 1\right)} + \frac{\operatorname{atan}{\left(x \right)}}{2}$$
Simplify
The graph
Plot the graph f(x)
The answer [src] 
1   pi
- + --
4   8
$$\frac{1}{4} + \frac{\pi}{8}$$ 
=
= 
1   pi
- + --
4   8
$$\frac{1}{4} + \frac{\pi}{8}$$ 
1/4 + pi/8
Numerical answer [src] 
0.642699081698724
0.642699081698724
The graph
Integral of 1/(x²+1)² dx

    Use the examples entering the upper and lower limits of integration.

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