Question

求不定积分∫根号(9+x^2)dx

求不定积分∫根号(9+x^2)dx

Answer 1

分部积分法：
∫√(9+x^2)dx
＝x√(9+x^2)dx－∫x^2/√(9+x^2)dx
＝x√(9+x^2)dx－∫(9+x^2-9)/√(9+x^2)dx
＝x√(9+x^2)dx－∫√(9+x^2)dx＋9∫1/√(9+x^2)dx
＝x√(9+x^2)dx－∫√(9+x^2)dx＋9ln(x+√(9+x^2))
∴∫√(9+x^2)dx＝1/2×x√(9+x^2)dx＋9/2×ln(x+√(9+x^2))＋C

Answer 2

所求（分部积分）=x√(9+x^2)-∫x^2/√(9+x^2)dx=x√(9+x^2)-所求+∫9/√(9+x^2)dx
而∫9/√9+x^2dx=ln(x/3+√(1+x^2/9)) (三角换元，或当公式记：∫1/√(x^2+1)dx=ln(x+√x^2+1)+c)
所以 所求为1/2(x√(9+x^2)+ln(x/3+√(1+x^2/9))+c

Answer 3

令x=3tant
dx=d3tant
∫√(9+x^2)dx
＝∫3sectd3tant

Answer 4

your answer here

Answer 5

错了，没看到根号