高数,不定积分题。
\u4e0d\u5b9a\u79ef\u5206 \u9ad8\u6570\u9898\u5982\u56fe\uff0c\u7136\u540e\u628at\u6362\u6389\u5c31\u597d\u4e86
arctanx\u7684\u539f\u51fd\u6570\u7684\u8ba1\u7b97\uff1a
\u222barctanxdx = xarctanx - \u222bxdarctanx + C
= xarctanx - \u222bx/(1+x²)dx + C
= xarctanx - ½\u222bd(1+x²)/(1+x²) + C
= xarctanx - ½ln(1+x²) + C
\u222barctanxdx (\u79ef\u5206\u533a\u95f4\uff1aa\u2192b)
=[xarctanx - ½ln(1+x²)]|(a\u2192b)
=b arctanb - a arctana - ½ln[(1+b²)/(1+a²)]
d/dx[\u222barctanxdx (\u79ef\u5206\u533a\u95f4\uff1aa\u2192b)] = 0
\u53ef\u4ee5\u6839\u636e\u5b9a\u79ef\u5206\u662f\u5e38\u6570\uff0c\u5e38\u719f\u7684\u5bfc\u6570\u662f0\uff0c\u5f97\u4ee5\u5224\u65ad\u3002
详细解答见图:
(11).(8),(16),(2)都用代换法;
(11)设z=1+2t;则dz=2dt; 原式=2∫(1/z)dz=2ln|t+C|=2ln|1+2t|+C;
(8)设t=1/x;则dt=- (1/x²)dx; 那么原式= - ∫e^tdt=-e^t+C=-e^(1/x)+C;
(16)设t=e^x;则dt=e^xdx; 那么原式=∫costdt=sint+C=sine^x +C;
(1)原式=(⅓x³-x)6₂=66-2/3=65+⅓;
(2)原式=(¼x⁴-x³)¹-₁=-2;
(2)设t=2y-3;则dt=2dy; 原式=½∫(1/t²)dt=- ½(1/t)+C=(1/2)(1/2y-3)+C
11.∫[1/(1+2t)]dt=(1/2)∫[1/(1+2t)]d(1+2t)=(1/2)ln︱1+2t︱+c
8. =-∫e^(1/x)d(1/x)=-e^(1/x)+c
16. =∫cose^xde^x=sine^x+c
1. =[(1/3)x^3-x](2,6)=[(1/3)6^3-6]-[(1/3)2^3-2]=196/3
2. =2∫(0,1)(-6)x^2dx=2*(-6)*(1/3)=-4
2. =-(1/2)[1/(2y-3)]+c=-1/2(2y-3)+c
前面几个都是凑微分,大同小异啊,我就说第一个吧
∫dt/1 2t=1/2∫d(2t 1)/2t 1
=1/2ln(2t 1) c
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