# evaluate the definite integral [0,1] of ln(4+x^2)dx

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should use integration by parts such that:

`int udv = uv -  int vdu`

Considering `u=ln(4+x^2)`  and `dv = dx`  yields:

`u=ln(4+x^2) => du = (2x)/(4+x^2)`

`dv = dx => v= x`

Using the formula of integration by parts yields:

`int ln(4+x^2) dx = xln(4+x^2)- int (2x^2)/(4+x^2)dx`

`int ln(4+x^2) dx = xln(4+x^2)- 2int (x^2)/(4+x^2)dx`

You need to evaluate `int (x^2)/(4+x^2)dx` , hence, you may add and subtract 4 to numerator such that:

`int (x^2 + 4 - 4)/(4+x^2)dx = int (x^2 + 4)/(4+x^2)dx - 4 int 1/(x^2+4)dx`

`int (x^2 + 4 - 4)/(4+x^2)dx = int dx -4 int 1/(x^2+2^2)dx`

`int (x^2 + 4 - 4)/(4+x^2)dx = x - 4*(1/2) arctan (x/2) + c`

`int (x^2 + 4 - 4)/(4+x^2)dx = x - 2arctan (x/2) + c`

`2int (x^2 + 4 - 4)/(4+x^2)dx = 2x - 4arctan (x/2) + c`

`int ln(4+x^2) dx = xln(4+x^2) - 2x+ 4arctan (x/2) + c`

You may evaluate the definite integral using the fundamental theorem of calculus such that:

`int_0^1 ln(4+x^2) dx = (xln(4+x^2) - 2x + 4arctan (x/2))|_0^1`

`int_0^1 ln(4+x^2) dx = (ln(4+1^2) - 2 + 4arctan (1/2) - 0 + 0 - 0)`

`int_0^1 ln(4+x^2) dx = (ln5 - 2 + 4arctan (1/2))`

Hence, evaluating the given definite integral yields `int_0^1 ln(4+x^2) dx = (ln5 - 2 + 4arctan (1/2)).`