How do you integrate #x/((ln3)(x^2 + 4))#? Calculus Techniques of Integration Integration by Parts 1 Answer Antoine May 2, 2015 #intx/((ln3)(x^2+4))dx# #ln3# is a constant so this turns down to, #=> 1/ln3intx/(x^2+4)dx =1/ln3 int1/2*(2x)/(x^2+4)dx=1/(2ln3)int(2x)/(x^2+4)dx# The numerator is the derivative of the denominator so, #=>1/(2ln3)lnabs(x^2+4) + C# Answer link Related questions How do I find the integral #int(x*ln(x))dx# ? How do I find the integral #int(cos(x)/e^x)dx# ? How do I find the integral #int(x*cos(5x))dx# ? How do I find the integral #int(x*e^-x)dx# ? How do I find the integral #int(x^2*sin(pix))dx# ? How do I find the integral #intln(2x+1)dx# ? How do I find the integral #intsin^-1(x)dx# ? How do I find the integral #intarctan(4x)dx# ? How do I find the integral #intx^5*ln(x)dx# ? How do I find the integral #intx*2^xdx# ? See all questions in Integration by Parts Impact of this question 1532 views around the world You can reuse this answer Creative Commons License