Find the integral of 1 1 x 4 .

Points of 1 (1+x 4):

1/(1+x^4) dx

1/2)∫ 1-x)+(1+x)]/(1+x^4) dx

1/2)∫ 1-x)/(1+x^4) dx + 1/2)∫ 1+x)/(1+x^4) dx

Formula for indefinite integrals:

1. A dx = ax + c, a and c are constants.

2. x a dx = [x (a + 1)] (a + 1) +c, where a is a constant and a ≠ 1

3、∫ 1/x dx = ln|x| +c

4. A x DX = (1 LNA)a x + C, where A > 0 and A ≠ 1

5、∫ e^x dx = e^x + c

6、∫ cosx dx = sinx + c

7、∫ sinx dx = - cosx + c

8、∫ cotx dx = ln|sinx| +c = - ln|cscx| +c

1+x^41+x²)²

2x²1+x²+√2x)(1+x²-√2x)1/(1+x^4)

1/(1+x²-√2x)

1/(1+x²+√2x)]/2√2x

1/x√2-x)/(1+x²-√2x)

1/x√2+x)/(1+x²+√2x)]

2x+2√2)/(x²+√2x+1)

2x-2√2)/(x²+1-√2x)]

2x+√2)/(x²+√2x+1)

2x-√2)/(x²+1-√2x)

2/(x²+√2x+1)

2/(x²+1-√2x)]

Finding the integral for (2x+ 2) (x + 2x+1) gives ln(x + 2x+1).

Finding the integral for (2x- 2) (x +1- 2x) yields ln(x +1- 2x).

Integrals 2 (x + 2x+1) to get 2arctan( 2x+1) and 2 (x - 2x+1) to get 2arctan( 2x-1) primitive. c

1/(1+x⁴)dx

1+x²+1-x²)/(1+x⁴)dx

1+x²)/(1+x⁴)dx

1-x²)/(1+x⁴)

The numerator and denominator of DX are divided by X

1/x²1)/(1/x²x²)dx

1/x²)/(1/x²x²)dx

The molecule is placed after the differential symbol.

1/(1/x²x²

d(x-1/x)

1/(1/x²x²

d(x+1/x)

1/[(x-1/x)²

d(x-1/x)

1/[(x1/x)²

d(x+1/x)

2/4)arctan[(x-1/x)/√2]√2/8)ln|(x

1/x2)/(x1/x

c(√2/4)arctan[(x-1/x)/√2]√2/8)ln|(x²

2x)/(x²

2x)c The Beauty of Mathematics] team will answer for you, if you don't understand, please ask, if you solve the problem, please click "Choose as a satisfactory answer" below.

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dx lead per mu x (1 + x 4) = x 3 x raid x 4 (1 + x 4) = dx 4 4 (x 4 + x 8) = dx 4 4x 4 + dx 4 4 (1 + x 4) = (ln x 4) 4-ln(1 + x 4) 4

Up and down x 3, it's very clear Xun Zen is good, write it down and see it.

The practice of using Chang to do the previous one is large, and it can be divided into parts, but it is very complex and resistant.

dx (1+x 4) =1 2)[ 1+x )dx (1+x 4)+ 1-x )dx (1+x 4)] Both integrals are numerator denominators divided by x ) 1 2) =1 2) =1 2)+c =[1 (2 2)]arctan[(x -1) x 2] -1 4 2)ln[(x -x 2+1) (x +x 2+1)]+c.

If the posture is higher than this integral, first see if there is a solution to the denominator x 4+1=0, like there is no real number solution in this problem So the denominator must be able to represent the form of x 4+1=(x 2+ax+b)(x 2+cx+d) with the undetermined coefficient method to obtain a= 2,c=- 2,b=d=1 and then according to the method of dismantling the fraction of the calendar stockings The original formula must be expressed as: 1 (x 4+1)=(ax+..

Here's how, please refer to:

<> a bit of a troublesome problem, mainly the splitting of polynomials.