
Yefim S. answered 03/17/24
Math Tutor with Experience
x = 5/2tanu; dx = 5/2sec2udu; ∫x2/(4x2 + 25)3/2dx = ∫25/4tan2u/(53sec3u)·5/2sec2udu = 1/8∫tan2u/secudu =
1/8∫(secu - cosu)du = 1/8[ln(secu + tanu) - sinu] + C = 1/8[ln(2/5x + √(1 + 4/25x2)) - 2/5x/√(1 + 4/25x2)] + C