求不定积分∫根号(9+x^2)dx

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...

您好，答案如图所示：

--> x = 3 siny，dx = 3 cosy dy --> = ∫ √(9 - 9sin²y) 3cosy dy = ∫ 3cosy 3cosy dy = 9∫ cos²y dy = (9/2)∫ (1 + cos2y) dy = (9/2)[y + (1/2)sin2y] + C = (9/2)y + (9/2)sinycosy + C = (9/2)arcsin(x/3) + (9/2...

可以采用三角变换求解。具体做法是令x＝3sint 那么x^2＝9sin^2t dx＝3costdt 原式子变为 ∫（9－9sin^t）^（1/2）×3costdt ＝9∫cos^2td（利用算法1－cos^t＝sin^t）＝9/2∫（cos2t＋1）dt（利用算法cos2t＝2cos^2t－1）＝9/4sin2t＋（9/2）t＋c，c为任一常数。然后再把t...

∫√(9-x^2)dx=(9/2)arcsin(x/3)+(1/2)x)√[(9-(x²)]+C，(C为任意常数)。令x=3sint，则dx=3costdt.t=arcsin(x/3)，sin2t=2sintcost。∫√(9-x^2)dx =∫[√(9-9sin²t)]3(cost)dt =∫9cos²tdt=9∫(1/2)[1+cos(2t)]dt =9∫(1/4)[1+...

设x=3sint ∫x^2/根号下(9-X^2) dx =∫9(sint)^2*3costdt/3cost =(9/2)∫(1-cos2t)dt =(9/4)∫(1-cos2t)d(2t)=(9/4)(2t-sin2t)+C =(9/2)(t-sintcost)+C =(9/2)[arcsin(x/3)-x/3*根号(9-x^2)]+C ...

先做不定积分 令x=3tanu，则√(x^2+9)=3secu，dx=3sec²udu，∫ √(x^2+9)dx =∫ 3secu*3sec²udu =9∫ sec³udu 下面计算：∫ sec³udu =∫ secud(tanu)=secutanu-∫ tan²usecudu =secutanu-∫ (sec²u-1)secudu =secutanu-∫ sec³...

或直接用积分表公式：∫ dx/√(x² + a²) = ln|x + √(x² + a²)| + C 则∫ dx/√(9x² + 1) = ∫ dx/[3√(x² + 1/9)]= (1/3)∫ dx/√(x² + (1/3)²) = (1/3)ln|x + √(x² + (1/3)²)|...

解：令x=3cosα，则dx=-3sinα dα ∫1/x√(9-x²) dx =∫1/[3cosα·3sinα](-3sinα) dα =-1/3 ∫ 1/(cosα) dα =-1/3 ∫secα dα =-1/3 ln|secx+tanx|+C =-13·ln|3/x+√(9/x²-1)|+C ...

4,令x=2sect,原式=1/2t,再带换回x即可.5,过程比较难表达.方法是令x=atant,令分母变为asect,会比较好算一些了.,4,求不定积分 2.∫(1+tanx)/cos^2x dx 3.∫x^2/√（9-x^2) dx 4.∫1/x√(x^2-4) dx 5.∫dx/√(a^2+x^2)要思路、解法、步骤,结果我有,