∫yln(1+xy)dy上限是1/x,下限是0∫yln(1+xy)dy上限是1/x,下限是0

Question

∫yln(1+xy)dy 上限是1/x ,下限是0
∫yln(1+xy)dy 上限是1/x ,下限是0

Answer 1

∫(0→1/x) yln(1 + xy) dy
u = 1 + xy,du = x dy
y = 0,u = 1；y = 1/x,u = 2
原式 = ∫(1→2) (u - 1)/x * ln(u) * du/x
= (1/x²)∫(1→2) [uln(u) - ln(u)] du
= (1/x²)∫(1→2) ln(u) d(u²/2) - (1/x²)[uln(u) - u] |[1→2]
= (1/2x²)*u²ln(u) |[1→2] - (1/2x²)∫(1→2) u² * 1/u du - (1/x²){[2ln(2) - 2] - (- 1)}
= (1/2x²)*4ln(2) - (1/4x²)*u² |[1→2] - (1/x²)[2ln(2) - 1]
= [2ln(2)]/x² - 3/(4x²) - [2ln(2)]/x² + 1/x²
= - 3/(4x²) + 4/(4x²)
= 1/(4x²)