∫x²e^(2x)dx怎么写 ∫(e^t)f(x-t)dt=x(e^2x),求∫f(x)d...
\u222bxe^x/\u221a(1+e^x)dx2x\u221a(1 + e^x) - 4\u221a(1 + e^x) + 2ln| [\u221a(1 + e^x) + 1]/[\u221a(1 + e^x) - 1] | + C
\u89e3\u9898\u8fc7\u7a0b\u5982\u4e0b\uff1a
\u8bbet = \u221a(1 + e^x),x = ln(t² - 1),dx = 2t/(t² - 1) dt
\u222b xe^x/\u221a(1 + e^x) dx
= \u222b [ln(t² - 1) * (t² - 1)/t] * 2t/(t² - 1) dt
= 2\u222b ln(t² - 1) dt
= 2t ln(t² - 1) - 2\u222b t d[ln(t² - 1)]
= 2t ln(t² - 1) - 2\u222b t * 2t/(t² - 1) dt
= 2t ln(t² - 1) - 4\u222b [(t² - 1) + 1]/(t² - 1) dt
= 2t ln(t² - 1) - 4\u222b dt - 4\u222b 1/(t² - 1) dt
= 2t ln(t² - 1) - 4t - 4(1/2)\u222b [1/(t - 1) - 1/(t + 1)] dt
= 2t ln(t² - 1) - 4t - 2ln|t - 1| + 2ln|t + 1| + C
= 2x\u221a(1 + e^x) - 4\u221a(1 + e^x) + 2ln| [\u221a(1 + e^x) + 1]/[\u221a(1 + e^x) - 1] | + C
\u8bb0\u4f5c\u222bf(x)dx\u6216\u8005\u222bf\uff08\u9ad8\u7b49\u5fae\u79ef\u5206\u4e2d\u5e38\u7701\u53bbdx\uff09\uff0c\u5373\u222bf(x)dx=F(x)+C\u3002\u5176\u4e2d\u222b\u53eb\u505a\u79ef\u5206\u53f7\uff0cf(x)\u53eb\u505a\u88ab\u79ef\u51fd\u6570\uff0cx\u53eb\u505a\u79ef\u5206\u53d8\u91cf\uff0cf(x)dx\u53eb\u505a\u88ab\u79ef\u5f0f\uff0cC\u53eb\u505a\u79ef\u5206\u5e38\u6570\u6216\u79ef\u5206\u5e38\u91cf\uff0c\u6c42\u5df2\u77e5\u51fd\u6570\u7684\u4e0d\u5b9a\u79ef\u5206\u7684\u8fc7\u7a0b\u53eb\u505a\u5bf9\u8fd9\u4e2a\u51fd\u6570\u8fdb\u884c\u4e0d\u5b9a\u79ef\u5206\u3002
\u6269\u5c55\u8d44\u6599\uff1a\u5e38\u7528\u79ef\u5206\u516c\u5f0f\uff1a
1\uff09\u222b0dx=c
2\uff09\u222bx^udx=(x^(u+1))/(u+1)+c
3\uff09\u222b1/xdx=ln|x|+c
4\uff09\u222ba^xdx=(a^x)/lna+c
5\uff09\u222be^xdx=e^x+c
6\uff09\u222bsinxdx=-cosx+c
\u222b(0,x)f(t-x)dt=e^(-x²)+1 \u4ee4u=t-x 0<=t<=x\u5219-x<=u<=0 \u539f\u5f0f=\u222b(-x,0)f(u)d(u+x)=e^(-x²)+1 \u8fd9\u91ccx\u770b\u505a\u5df2\u77e5\u5e38\u6570 -\u222b(0,-x)f(u)du=e^(-x²)+1 \u4e24\u8fb9\u6c42\u5bfc -f(-x)*(-x)'=(e^(-x²)+1)'=e^(-x²)*(-x²)' f(-x)=-2xe^(-x²) f(x)=2xe^(-x²)
分部积分法。原式 = ∫1/2*x^2d(e^(2x)) = 1/2*x^2*e^(2x) - ∫x*e^(2x)dx
= 1/2*x^2*e^(2x) - ∫1/2*x d(e^(2x))
= 1/2*x^2*e^(2x) - 1/2*x*e^(2x) + 1/2*∫e^(2x)dx
= 1/2*x^2*e^(2x) - 1/2*x*e^(2x) + 1/4*e^(2x) + C
= 1/4*e^(2x)*(2x^2-2x+1) + C 。
绛旓細瑙ｇ瓟杩囩▼濡備笅锛氣埆 ln (x) dx =x ln (x) -鈭 x d [ ln(x) ]=x ln(x) -鈭 x *(1/x) dx =x ln (x) -鈭 dx =x ln (x) -x +C,(C涓轰换鎰忓父鏁).
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绛旓細鈭 xsinx dx=-xcosx+sinx+C銆傦紙C涓虹Н鍒嗗父鏁帮級瑙ｇ瓟杩囩▼濡備笅锛氬垎閮ㄧН鍒嗘硶锛氣埆udv=uv-鈭玽du 鈭 xsinx dx = - 鈭 x d(cosx)=-xcosx+鈭 cosx dx =-xcosx+sinx+C 涓嶅畾绉垎锛氫笉瀹氱Н鍒嗙殑绉垎鍏紡涓昏鏈夊涓嬪嚑绫伙細鍚玜x+b鐨勭Н鍒嗐佸惈鈭氾紙a+bx锛夌殑绉垎銆佸惈鏈墄^2卤伪锛2鐨勭Н鍒嗐佸惈鏈塧x^2+b...
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