Indefinite integral, where did the question 1 3 come from?

x/(1+x∧3)dx

Let x (1+x 3)=x [(1+x)(1-x+x 2)]=a (1+x)+(bx+c) (1-x+x 2).

The solution yields a=-1 3, b=1 3, and c=1 3

Original = [1 3) (1+x)+(1 3)(x+1) (1-x+x 2)]dx

1/3)ln|1+x|+(1/3)∫(1+x)/(1-x+x^2)dx

where (1+x) (1-x+x 2)dx

1/2)∫[2x-1)/(1-x+x^2)+3/(1-x+x^2)]dx

1/2)∫1/(1-x+x^2)d(1-x+x^2)+(3/2)∫1/(1-x+x^2)dx

1/2)ln(1-x+x^2)+(3/2)∫1/[(x-1/2)^2+3/4]dx

1/2)ln(1-x+x^2)+2∫1/dx

1 2)ln(1-x+x 2)+2* 3 2 1 d[(2 this elimination 3)(x-1 2)].

1 2) LN(1-x+x 2)+3arctan[(type 2 or 3)(x-1 2)]+c

x/(1+x∧3)dx

1/3)ln|1+x|+(1 6)ln(1-x+x 2)+(3 3)arctan[(2 3)(x-1 Bu Pavu2)]+c

1/6)ln[(1-x+x^2)/(1+x)^2]+(3/3)arctan[(2x-1)/√3)]+c

1 3 times under the root number (2-3x) dx

2-3x) (-1 to the 3rd power] dx

2-3x) to the power (-1 3)] *1 3) d(2-3x) (-1 3) * (3 2) * (2-3x) + (2 3) to the power (2) 3 + c (-1 2) * (2-3x).

Summary. For x, 82 2 is for t and is the (penultimate one) 1 (3 x+1) indefinite integral.

Teacher, why is this question equal to 3ln3?

I've figured this out, but why is the answer to that definite integral 3ln3?

Pro, the indefinite integral is calculated as the first graph.

Bring in the upper and lower limits and you get 3ln3 <>

Generally, this kind of irrational integral (the root number integral is to directly make a t=irrational commutation, and then replace it with a rational integral).

The kind of square term under the root number is generally the triangular exchange <>< used

However, the definite integral is not a limit for the exchange of yuan.

But I'm calculating indefinite integrals<>

So it's not 2 2 no.

Your problem is to find indefinite integrals, and it is the same to calculate indefinite integrals and bring them into the upper and lower bounds.

Looking at the final result, it would be good to bring in the upper limit 8 and 2.

It is possible to bring in x, to bring in 82 and 2 to bring in to t (penultimate one). The generation t is the generation t = 2, not the 2 2 that the relatives say, because t = 3 x3 8 = 2

Hello, this question should be more troublesome.

Let t=(1+x) root number 2

d(x) = root number 2 * dt

Denominator = (root number 2 * t) 2 + (root number 2) 2 denominator is squared out of the root number 2, and the numerator is the root number 2

You get 1 root number 2

The method is shown in the figure below, please check it carefully, and I wish you a happy study: