高等数学微积分题目,要尽量详细的解题过程!以及回答一下照片中题目下方的问题
2个回答
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第一个,
原式=d/dx{x∫[x^2,2x] ln(1+t^2)dt}+d/dx{∫[x^2,2x] tln(1+t^2)dt}
=∫[x^2,2x] ln(1+t^2)dt+[xln(1+4x^2)-xln(1+x^4)]+2xln(1+4x^2)-x^2ln(1+x^4) (x=1代入)
=∫[1,2] ln(1+t^2)dt+ln5-ln2+2ln5-ln2
=tln(1+t^2)[1,2]-∫[1,2] tdln(1+t^2)+3ln5-2ln2
=2ln5-ln2-∫[1,2] t*2t/(1+t^2)dt+3ln5-2ln2
=-2∫[1,2] [1-1/(1+t^2)]dt+5ln5-3ln2
=-2(t-arctant)[1,2]+5ln5-3ln2
=-2+2arctan2-π/2+5ln5-3ln2
第二个
f(x^3)*(x^3)-f(x^2)*(x^2)'
=3x^2f(x^3)-2xf(x^2)
原式=d/dx{x∫[x^2,2x] ln(1+t^2)dt}+d/dx{∫[x^2,2x] tln(1+t^2)dt}
=∫[x^2,2x] ln(1+t^2)dt+[xln(1+4x^2)-xln(1+x^4)]+2xln(1+4x^2)-x^2ln(1+x^4) (x=1代入)
=∫[1,2] ln(1+t^2)dt+ln5-ln2+2ln5-ln2
=tln(1+t^2)[1,2]-∫[1,2] tdln(1+t^2)+3ln5-2ln2
=2ln5-ln2-∫[1,2] t*2t/(1+t^2)dt+3ln5-2ln2
=-2∫[1,2] [1-1/(1+t^2)]dt+5ln5-3ln2
=-2(t-arctant)[1,2]+5ln5-3ln2
=-2+2arctan2-π/2+5ln5-3ln2
第二个
f(x^3)*(x^3)-f(x^2)*(x^2)'
=3x^2f(x^3)-2xf(x^2)
追问
答案是-12ln17,后面的ln部分是对了,但前面的arctan是哪出了问题吗?
追答
答案肯定错了
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