How do you integrate #int x arctan x # using integration by parts? Calculus Techniques of Integration Integration by Parts 1 Answer Bio Nov 4, 2015 #intxtan^{-1}(x)dx=1/2[(x^2+1)tan^{-1}(x)-x]+c#, where #c# is the constant of integration. Explanation: #intxtan^{-1}(x)dx=1/2intfrac{d}{dx}(x^2)tan^{-1}(x)dx# #=1/2[x^2tan^{-1}(x)-intx^2frac{d}{dx}(tan^{-1}(x))dx]# #=1/2x^2tan^{-1}(x)-1/2intx^2(frac{1}{1+x^2})dx# #=1/2x^2tan^{-1}(x)-1/2int(1-frac{1}{1+x^2})dx# #=1/2x^2tan^{-1}(x)-1/2(x-tan^{-1}(x))+c#, where #c# is the constant of integration #=1/2[(x^2+1)tan^{-1}(x)-x]+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 1507 views around the world You can reuse this answer Creative Commons License