不定积分题目，求详细过程。。。
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\u222bxcos(x²+1)dx
=1/2\u222bcos(x²+1)d(x²+1)
=1/2\u222bcosudu
=1/2sinu+C
=1/2sin(x²+1)+C
x=t^2
dx=2tdt
原式=∫2^t *2tdt
=2∫t 2^t dt
=2/ln2∫t d2^t
=2t 2^t /ln2 -2/ln2 ∫2^t dt
=2t*2^t /ln2 -2*2^t/(ln2)^2 +C
=t*2^(t+1)/ln2 -2^(t+1)/(ln2)^2 +C
=√x *2^(√x +1)/ln2 -2^(√x +1) /(ln2)^2 +C
绛旓細鏂规硶濡備笅锛岃浣滃弬鑰冿細鑻ユ湁甯姪锛岃閲囩撼銆
绛旓細(1)鈭2x.e^xdx =2鈭玿.de^x =2x.e^x -2鈭玡^x dx =2x.e^x -2e^x + C (2)鈭2x.sinxdx =-2鈭玿dcosx =-2xcosx +2鈭玞osxdx =-2xcosx +2sinx + C (3)鈭(x+1)^3 dx =鈭(x+1)^3 d(x+1)=(1/4)(x+1)^4 + C (4)鈭 4dx/[1+鈭(2x) ]let 鈭(2x) = (...
绛旓細1.鈭(5x-4)^4 dx =1/5 鈭(5x-4)^4d(5x-4)=1/25 (5x-4)^5 +C 3.鈭玿cosx²dx =1/2 鈭玞osx²d(x²)=1/2 sinx² +C 5.鈭玿²e^x³dx =1/3 鈭玡^x³d(x³)=1/3 e^x³ +C 7.鈭3^(1/x) /x² d...
绛旓細锛1锛夊師寮=鈭玔x^(3/2)+1-x-x^(-1/2)]dx =(2/5)*x^(5/2)+x-(1/2)*x^2-2x^(1/2)+C 锛2锛夊師寮=(1/ln10)*10^x+3sinx+2x^(1/2)+C 锛3锛夊師寮=鈭玔2/鈭(1-x^2)-1]dx =2arcsinx-x+C 锛4锛夊師寮=鈭玠x/2cos^2x =(1/2)*鈭玸ec^2xdx =(1/2)*tanx+C...
绛旓細鏂规硶濡備笅锛岃浣滃弬鑰冿細
绛旓細鍏蜂綋鍥炵瓟濡傚浘锛氳繛缁嚱鏁帮紝涓瀹氬瓨鍦ㄥ畾绉垎鍜涓嶅畾绉垎锛涜嫢鍦ㄦ湁闄愬尯闂碵a,b]涓婂彧鏈夋湁闄愪釜闂存柇鐐逛笖鍑芥暟鏈夌晫锛屽垯瀹氱Н鍒嗗瓨鍦紱鑻ユ湁璺宠穬銆佸彲鍘汇佹棤绌烽棿鏂偣锛屽垯鍘熷嚱鏁涓瀹氫笉瀛樺湪锛屽嵆涓嶅畾绉垎涓瀹氫笉瀛樺湪銆
绛旓細=sinxe^2x+2cosxe^2x-2鈭玞osxde^2x =sinxe^2x+2cosxe^2x+4鈭玞osxe^2xdx 鈭玞osxe^2x=-锛坰inxe^2x+2cosxe^2x锛/3 姹傚嚱鏁癴(x)鐨勪笉瀹氱Н鍒嗭紝灏辨槸瑕佹眰鍑篺(x)鐨勬墍鏈鐨勫師鍑芥暟锛鐢鍘熷嚱鏁扮殑鎬ц川鍙煡锛屽彧瑕佹眰鍑哄嚱鏁癴(x)鐨勪竴涓師鍑芥暟锛屽啀鍔犱笂浠绘剰鐨勫父鏁癈灏卞緱鍒板嚱鏁癴(x)鐨勪笉瀹氱Н鍒嗐
绛旓細杩囩▼濡備笅锛氣埆3e^(3x)dx =鈭玡^(3x)d锛3x锛=e^(3x)+C 杩炵画鍑芥暟锛屼竴瀹氬瓨鍦ㄥ畾绉垎鍜涓嶅畾绉垎锛涜嫢鍦ㄦ湁闄愬尯闂碵a,b]涓婂彧鏈夋湁闄愪釜闂存柇鐐逛笖鍑芥暟鏈夌晫锛屽垯瀹氱Н鍒嗗瓨鍦紱鑻ユ湁璺宠穬銆佸彲鍘汇佹棤绌烽棿鏂偣锛屽垯鍘熷嚱鏁涓瀹氫笉瀛樺湪锛屽嵆涓嶅畾绉垎涓瀹氫笉瀛樺湪銆
绛旓細浠 = 鈭歺鍒檟 = u²锛宒x = 2u du 鈭 a^鈭歺 dx = 2鈭 ua^u du = 2鈭 u d(a^u/lna)= (2/lna) ua^u - (2/lna)鈭 a^u du = (2/lna) ua^u - (2/lna) * a^u/lna + C = (2ua^u)/lna - (2a^u)/(lna)² + C = (2鈭歺 a^鈭歺)/lna - ...
绛旓細1銆佺绫绘崲鍏冩硶 鈭1/(1+e^x)dx锛濃埆e^(-x)/(1+e^(-x))dx锛-鈭1/(1+e^(-x))d(1+e^(-x))锛-ln(1+e^(-x))锛婥锛-ln((1+e^x)/e^x)锛婥锛漻-ln(1+e^x)锛婥 鎴 鈭1/(1+e^x)dx锛濃埆 [1 - e^x/(1+e^x))dx锛漻-鈭1/(1+e^x)d(1+e^x)锛漻-ln(1+...
