2018年4月8日 · (x-1)^4 -= x^4 -4 x^3 + 6x^2- 4x+1 We can expand the expression using the binomial theorem: (x-1)^4 -= sum_(r=0)^4 ( (n), (r) ) (x)^r(-1)^(n-r) " " = ( (4), (0) ) (x ...

2017年10月17日 · x=-2/5 or -0.4 Move 1 to the right hand side of the equation so you get rid of it. 1/(1+(1)/((1+1/x))=4-1 ...

2018年5月2日 · x^4+4x^3+6x^2+4x+1 The binomial theorem states: (a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 so here, a=x and b=1 We get: (x+1)^4 = …

2016年9月24日 · 64668 views around the world You can reuse this answer ...

2011年9月18日 · 1/(x^4+1) I think it also have been solved multiple times on this site. Could anyone help me find a complete solution? (Link to thread here) The basic outline I remember …

2017年10月3日 · The answer is long, so please see the below. Key points are factorization, finding coefficients of an identity, integration of (f'(x))/(f(x)) and 1/((x+p)^2+q). x^4+1 is not able …

1 - 4x +10x^2 - 20x^3 + 35x^4 +..... >rewrite as (1+x)^-4 Since there is a negative exponent use the following version of the binomial expansion.

2017年9月18日 · I = int_(-oo)^(oo) 1/(1+x^4) \\ dx = (pisqrt(2))/2 Sorry that this is such a long solution: We seek: I = int_(-oo)^(oo) 1/(1+x^4) \\ dx graph{1/(1+x^4) [-5, 5, -2.5, 2.5]} The …

2018年4月18日 · x^4-1=(x^2+1)(x+1)(x-1) using complex numbers x^4-1=(x+ib)(x-ib)(x+1)(x-1) we make use of the difference of squares a^2-b^2=(a+b)(a-b) x^4-1=(x^2+1)(x^2-1) we can use …

2015年5月5日 · int x/(1+x^4)dx =1/2arctan(x^2) Method int x/(1+x^4)dx = 1/2int (2xdx)/(1+(x^2)^2)=1/2int (d(x^2))/(1+(x^2)^2) ="I" let x^2 =u =>"I"= 1/2int (du)/(1+u^2) = …