不定积分求解，大神们先谢了。答案是什么？ 不定积分计算，求大神解答
\u6c42\u4e0d\u5b9a\u79ef\u5206\uff0c\u8981\u6709\u8fc7\u7a0b\u548c\u7b54\u6848\u3002 \u5728\u6b64\u5148\u8c22\u8fc7\u5927\u795e\u4eec\u3002\u5982\u56fe\uff0c
\u4e0d\u53ef\u521d\u7b49\u79ef\u5206\u3002
\u5206\u6bcd\u53bb\u6389\u4e00\u4e2a\u221ax\u8fd8\u53ef\u4ee5\u505a
积分区间为[-5，5]，关于原点对称。积分项是奇函数，积分结果为偶函数。
原式=0
不需要真的去进行积分计算，只需运用定积分的性质解题。
绛旓細浠 = sinz锛宒x = cosz dz 鈭 x/鈭(1 - x²) 路 e^(arcsinx) dx = 鈭 sinz/cosz 路 e^z 路 cosz dz = 鈭 e^z 路 sinz dz = 鈭 e^z d(- cosz)= - (e^z)cosz + 鈭 cosz de^z = - (e^z)cosz + 鈭 e^z d(sinz)= - (e^z)cosz + (e^z)...
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绛旓細缁撴灉姣旇緝闅剧畻锛屽鏋滅畻瀹氱Н鍒鐨勮瘽锛屽彲浠ュ埄鐢ㄥ鍑芥暟鐨勫绉版
绛旓細(1)鈭玿(arctanx)^2 dx =(1/2) 鈭(arctanx)^2 dx^2 =(1/2)x^2.(arctanx)^2 - 鈭玔x^2.(arctanx)/(1+x^2)] dx =(1/2)x^2.(arctanx)^2 - 鈭玔1 - 1/(1+x^2)] (arctanx) dx =(1/2)x^2.(arctanx)^2 - 鈭玜rctanx dx +鈭 arctanx/(1+x^2...
绛旓細浠3鈭(x+1) =t 鍒檟=t^3 -1锛宒x=3t²dt 鍘熷紡=鈭3t²dt/(t+1)=3鈭(t²-1+1)dt/(t+1)=3鈭玔t-1+1/(t+1)]dt =3[t²/2-t+ln|t+1|]+C =3t²/2-3t+3ln|t+1|+C =3/2 (x+1)^(2/3) - 3(x+1)^(1/3) + 3ln|(x+1)^...
绛旓細棣栧厛姹傗埆sec^3(x) dx锛氳I=鈭玸ec^3(x) dx锛屽垯I =鈭玸ec(x)*sec^2(x) dx =鈭玸ec(x)*[tan(x)]' dx =sec(x)*tan(x)-鈭玔sec(x)]'*tan(x) dx =sec(x)*tan(x)-鈭玔sec(x)*tan(x)]*tan(x) dx =sec(x)*tan(x)-鈭玸ec(x)*tan^2(x) dx =sec(x)*tan(x)-鈭...
绛旓細绗竴棰=鈭(X^2+X^(-2)/X^3DX=鈭玐^(-1)+X^(-5)DX=ln|x|-4/x^(-5)+c 绗簩棰樿arcsinx=t锛寈=sint 鈭(arcsinx)^2dx=鈭玹^2costdt=t^2sint-鈭2tsintdt+c=t^2sint+2tcost-鈭2costdt+c=t^2sint+2tcost-2sint+c 绗笁棰樹袱杈规眰瀵煎緱sinx/f(x)=-sinx/(1+cos²x)...
绛旓細=(cos^2 x- sin^2 x)/[sin^2 x cos^2 x]=1/sin^2 x - 1/ cos^2 x 鍒嗗埆绉垎 =-cotx-tanx+C
绛旓細e^xdx/(1+cosx)+鈭玡^xsinxdx/(1+cosx)=鈭玡^xdtan(x/2)+鈭玹an(x/2)de^x =e^x tan(x/2) -鈭玹an(x/2)de^x+鈭玹an(x/2)de^x+C =e^x tan(x/2) +C
