Nitin P. answered • 07/01/20

B.S. in Computer Science and Applied Mathematics from UC Berkeley

Let x = (3/5)tan z. Then z = tan^{-1}(5x/3), dx = 3/5sec^{2}z dz and we have:

∫dx/(x(9+25x^{2})^{1/2}) = (3/5) ∫sec^{2}z dz/[(3/5)tan z(9 + 9tan^{2}z)^{1/2}]

= (1/3) ∫sec^{2}z dz/(tan z sec z)

= (1/3) ∫sec z dz/(sin z sec z)

= (1/3) ∫ csc z dz = -(1/3) ln |csc z + cot z| + C

= -(1/3) ln |csc(tan^{-1}(5x/3)) + cot (tan^{-1}(5x/3))| + C

= -(1/3) ln|sqrt(25x^{2} + 9)/(5x) + 3/(5x)| + C