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

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)).

Approved by eNotes Editorial Team

Posted on