① ∫(2x+4)/(x2 +2x+3) dx; ② ∫(x2)/(1+x2)arctanx dx; ③ 1/[(3√x

草民的忧虑2022-10-04 11:39:541条回答

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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sun720616 共回答了27个问题 | 采纳率96.3%
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.
1年前

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f(x)=arctanx
f'(x)=1/(1+x²)>0
函数是增函数
所以
f(0)=0.最小值
f(1)=arctan1=π/4
所以
0