赞 rimyo78 幼苗
共回答了21个问题采纳率：85.7% 举报
∫1/(x^4+1)dx
=1/2∫[(x²+1)-(x²-1)]/(x^4+1)dx
=1/2[∫(x²+1)/(x^4+1)dx-∫(x²-1)/(x^4+1)dx]
=1/2[∫(1+1/x²)/(x²+1/x²)dx-∫(1-1/x²)/(x²+1/x²)dx]
=1/2[∫1/(x²+1/x²-2+2)d(x-1/x)-∫1/(x²+1/x²+2-2)d(x+1/x)]
=1/2[∫1/[(x-1/x)²+2]d(x-1/x)-∫1/[(x+1/x)²-2]d(x+1/x)]
=1/2[√2/2arctan[√2/2(x-1/x)]-√2/4ln|(x²-√2x+1)/(x²+√2x+1)|]+C
=√2/4arctan[√2/2(x-1/x)]-√2/8ln|(x²-√2x+1)/(x²+√2x+1)|+C
=1/2∫[(x²+1)-(x²-1)]/(x^4+1)dx
=1/2[∫(x²+1)/(x^4+1)dx-∫(x²-1)/(x^4+1)dx]
=1/2[∫(1+1/x²)/(x²+1/x²)dx-∫(1-1/x²)/(x²+1/x²)dx]
=1/2[∫1/(x²+1/x²-2+2)d(x-1/x)-∫1/(x²+1/x²+2-2)d(x+1/x)]
=1/2[∫1/[(x-1/x)²+2]d(x-1/x)-∫1/[(x+1/x)²-2]d(x+1/x)]
=1/2[√2/2arctan[√2/2(x-1/x)]-√2/4ln|(x²-√2x+1)/(x²+√2x+1)|]+C
=√2/4arctan[√2/2(x-1/x)]-√2/8ln|(x²-√2x+1)/(x²+√2x+1)|+C
1年前
33
可能相似的问题
-
1年前2个回答
-
设函数f(x)=x的四次幂加(m加3)x的三次幂是偶函数,则m=
1年前1个回答
-
1年前3个回答
-
1年前4个回答
-
1年前1个回答
-
1年前1个回答
-
1年前2个回答
-
1年前4个回答
你能帮帮他们吗