How do you find the indefinite integral of int (4/(3t^2)+7/(2t))dt? Calculus Introduction to Integration Integrals of Polynomial functions 1 Answer Shwetank Mauria Feb 25, 2017 int(4/(3t^2)+7/(2t))dt=-4/(3t)+7/2lnt+c Explanation: As intt^ndt=t^(n+1)/(n+1)+c, but for n=-1, intt^(-1)dt=int(dt)/t=lnt int(4/(3t^2)+7/(2t))dt = int(4/3t^(-2)+7/2t^(-1))dt = 4/3t^(-1)/(-1)+7/2lnt+c = -4/(3t)+7/2lnt+c Answer link Related questions How do you evaluate the integral intx^3+4x^2+5 dx? How do you evaluate the integral int(1+x)^2 dx? How do you evaluate the integral int8x+3 dx? How do you evaluate the integral intx^10-6x^5+2x^3 dx? What is the integral of a constant? What is the antiderivative of the distance function? What is the integral of |x|? What is the integral of 3x? What is the integral of 4x^3? What is the integral of sqrt(1-x^2)? See all questions in Integrals of Polynomial functions Impact of this question 4973 views around the world You can reuse this answer Creative Commons License