x的6次方加一不定积分怎么算

Question

如题所述

Answer 1

您好：∫（x＾6＋1）dx=∫x＾6dx＋∫dx＝1／7 x＾7＋x＋c
祝学习愉快

Answer 2

>> int('1/(1+x^6)')
ans =
1/3*atan(x)-1/12*3^(1/2)*log(x^2-3^(1/2)*x+1)+1/6*atan(2*x-3^(1/2))+1/12*3^(1/2)*log(x^2+3^(1/2)*x+1)+1/6*atan(2*x+3^(1/2))

Answer 3

>> int('1/(1+x^6)')
ans =
1/3*atan(x)-1/12*3^(1/2)*log(x^2-3^(1/2)*x+1)+1/6*atan(2*x-3^(1/2))+1/12*3^(1/2)*log(x^2+3^(1/2)*x+1)+1/6*atan(2*x+3^(1/2))

Answer 4

x^6/(1+x²) =(x^6+1)/(1+x²)-1/(1+x²) =(x²+1)(x^4-x²+1)/(1+x²)-1/(1+x²) =x^4-x²+1-1/(1+x²) 所以原式=x^5/5-x³/3+x-arctanx+C