① ∫(2x+4)/(x2 +2x+3) dx; ② ∫(x2)/(1+x2)arctanx dx; ③ 1/[(3√x)+1] dx 注:注:x后的2为平方,根号前的3为开立方;
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① ∫(2x+4)/(x2 +2x+3) dx; ② ∫(x2)/(1+x2)arctanx dx; ③ 1/[(3√x)+1] dx 注:
注:x后的2为平方,根号前的3为开立方;
1、原式=∫d(x^2+2x+3)/(x^2+2x+3)+2∫dx/(x^2+2x+3)
=ln|x^2+2x+3|+2∫dx/[(x+1)^2+2]
=ln|x^2+2x+3|+√2∫d[(x+1)/√2]/{[(x+1)/√2]^2+1}
=ln|x^2+2x+3|+√2arctan(x+1)/√2+C.
2、原式=∫(1+x^2-1)arctanxdx/(1+x^2)
=∫arctanxdx-∫arctanxdx/(1+x^2)
=x*arctanx-∫xdx/(1+x^2)-∫arctanxd(arctanx)
=x*arctanx-(1/2)∫d(1+x^2)/(1+x^2)-(1/2)(arctanx)^2
=xarctanx-(1/2)ln(1+x^2)-(1/2)(arctanx)^2+C.
3、设x^(1/3)=t,
x=t^3,
dx=3t^2dt,
原式=∫3t^2dt/(t+1)
=3∫(t+1)dt-6∫dt+3∫d(t+1)/(t+1)
=3(t^2/2+t)-6t+3ln|1+t|+C
=3[x^(2/3)/2+x^(1/3)]-6x^(1/3)+3ln|1+x^(1/3)|+C
=(3/2)x^(2/3) -3x^(1/3)+3ln|1+x^(1/3)|+C.
(x2-3x)2-x(x2-3x)-2x2
通分 3x/x2-2x,2x+1/4-x2,3/2x2+4x
3/x2 -3x+2-1/x-2=1/x2 -x+4/x2 -2x
计算:x2-5x+6/x2-4x+3 + x2-4x+3/x2-2x+1 + x2-3x-4/1-x2
因式分解:(x2+2x+4)(x2+2x-3)-8
(x2+2x-3)(x2+2x+4)+6因式分解
求不等式(x2-2x-3)(x2-4x+3)
①x/(x2-1) ·(x2+x)/x2其中x=2 ②(x2-1)/(x2+4x+4)÷(x+1).(x+2﹚/﹙1-x﹚其中x=-3
5x2—[x2+(4x—x2)—2(x2—3x)]
1.(x2-4x+3)(x+2)
(x+1)(x2-x+1)+(x-2)(x2+2x+4),其中x=-3/2
分解因式:4(3x2-x-1)(x2+2x-3)-(4x2+x-4)2
(X2-X)2+(X2+3X+2)2-4(X2+X+1)2因式分解
先化简,再求值:(4x2-3x)-2(x2+2x-1)-(x2+x-1)
x2-2x-3
X2-2X-3
-x2+2x-3
(x-x2)(x2+x-6)分之(x2+3x)(x2-3x+2)