Mister Exam

Integral of ln(1+x²) dx

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

You have entered [src]
  1               
  /               
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 |     /     2\   
 |  log\1 + x / dx
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0                 
$$\int\limits_{0}^{1} \log{\left(x^{2} + 1 \right)}\, dx$$
Integral(log(1 + x^2), (x, 0, 1))
Detail solution
  1. Use integration by parts:

    Let and let .

    Then .

    To find :

    1. The integral of a constant is the constant times the variable of integration:

    Now evaluate the sub-integral.

  2. The integral of a constant times a function is the constant times the integral of the function:

    1. Rewrite the integrand:

    2. Integrate term-by-term:

      1. The integral of a constant is the constant times the variable of integration:

      1. The integral of a constant times a function is the constant times the integral of the function:

          PiecewiseRule(subfunctions=[(ArctanRule(a=1, b=1, c=1, context=1/(x**2 + 1), symbol=x), True), (ArccothRule(a=1, b=1, c=1, context=1/(x**2 + 1), symbol=x), False), (ArctanhRule(a=1, b=1, c=1, context=1/(x**2 + 1), symbol=x), False)], context=1/(x**2 + 1), symbol=x)

        So, the result is:

      The result is:

    So, the result is:

  3. Add the constant of integration:


The answer is:

The answer (Indefinite) [src]
  /                                                    
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 |    /     2\                                 /     2\
 | log\1 + x / dx = C - 2*x + 2*atan(x) + x*log\1 + x /
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$$\int \log{\left(x^{2} + 1 \right)}\, dx = C + x \log{\left(x^{2} + 1 \right)} - 2 x + 2 \operatorname{atan}{\left(x \right)}$$
The graph
The answer [src]
     pi         
-2 + -- + log(2)
     2          
$$-2 + \log{\left(2 \right)} + \frac{\pi}{2}$$
=
=
     pi         
-2 + -- + log(2)
     2          
$$-2 + \log{\left(2 \right)} + \frac{\pi}{2}$$
-2 + pi/2 + log(2)
Numerical answer [src]
0.263943507354842
0.263943507354842
The graph
Integral of ln(1+x²) dx

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