How do you determine the indefinite integral of (9/(4+z^2))dz? Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer Massimiliano May 9, 2015 In this way, remembering that int(f'(x))/(1+[f(x)]^2)dx=arctgf(x)+c: int(9/(4+z^2))dz=9int1/(4(1+z^2/4))dz=9/4int1/(1+(z/2)^2)dz= =2*9/4int(1/2)/(1+(z/2)^2)dz=9/2arctg(z/2)+c. Answer link Related questions How do you find the integral int1/(x^2*sqrt(x^2-9))dx ? How do you find the integral intx^3/(sqrt(x^2+9))dx ? How do you find the integral intx^3*sqrt(9-x^2)dx ? How do you find the integral intx^3/(sqrt(16-x^2))dx ? How do you find the integral intsqrt(x^2-1)/xdx ? How do you find the integral intsqrt(x^2-9)/x^3dx ? How do you find the integral intx/(sqrt(x^2+x+1))dx ? How do you find the integral intdt/(sqrt(t^2-6t+13)) ? How do you find the integral intx*sqrt(1-x^4)dx ? How do you prove the integral formula intdx/(sqrt(x^2+a^2)) = ln(x+sqrt(x^2+a^2))+ C ? See all questions in Integration by Trigonometric Substitution Impact of this question 1454 views around the world You can reuse this answer Creative Commons License