Question #178d6 Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Eddie Aug 31, 2016 #= 1/1000 ln ((x)/ (1000-x) )+ C # Explanation: #int 1/[x(1000-x)] dx # partial fractions is easiest/ most obvious way #=1/1000 int 1000/[x(1000-x)] dx # #= 1/1000 int (1000-x+x)/[x(1000-x)] dx # #= 1/1000 int (1000-x)/[x(1000-x)] + (x)/[x(1000-x)] dx # #= 1/1000 int (1)/[x] + (1)/(1000-x) dx # #= 1/1000 ( ln x - ln (1000-x) + C # #= 1/1000 ln ((x)/ (1000-x) )+ C # Answer link Related questions How do I evaluate the indefinite integral #intsin^3(x)*cos^2(x)dx# ? How do I evaluate the indefinite integral #intsin^6(x)*cos^3(x)dx# ? How do I evaluate the indefinite integral #intcos^5(x)dx# ? How do I evaluate the indefinite integral #intsin^2(2t)dt# ? How do I evaluate the indefinite integral #int(1+cos(x))^2dx# ? How do I evaluate the indefinite integral #intsec^2(x)*tan(x)dx# ? How do I evaluate the indefinite integral #intcot^5(x)*sin^4(x)dx# ? How do I evaluate the indefinite integral #inttan^2(x)dx# ? How do I evaluate the indefinite integral #int(tan^2(x)+tan^4(x))^2dx# ? How do I evaluate the indefinite integral #intx*sin(x)*tan(x)dx# ? See all questions in Integrals of Trigonometric Functions Impact of this question 1358 views around the world You can reuse this answer Creative Commons License