Mother in law is not a mother after all, why did Bao Wenjing give birth to a child sting tens of mil

Maybe it's really too humble to love, and it's not cared for.

The amniotic water broke, the mother-in-law was crying and worried about her daughter, but the mother-in-law was smiling and looking forward to the grandchildren, these two situations really reflected some shortcomings in society.

Because the scene of Bao Wenjing giving birth to a child is too similar to the treatment of most women in reality.

Because it's so hard for women, especially when it's too painful to have a baby.

Because Bao Wenjing's situation was more urgent at that time, her mother-in-law was not very worried about her.

I don't know, but maybe this magnifies the plight of all contemporary women.

If it's not her own mother, she will never really love herself, and as a mother-in-law, she looks forward to the next generation of her family.

Because her daughter is still distressed, her mother-in-law will never be as good as her mother.

I still don't know much about this, and I still don't know much about this.

I also remembered that when I gave birth, my mother-in-law and my eldest sister said that at this time, they were willing to see me uncomfortable, and if they were uncomfortable, they were about to give birth, and my mother wiped her tears next to me, but my mother-in-law and eldest sister were usually very good to me.

It doesn't hurt, it doesn't hurt, it's already numb.

Treating my mother-in-law as a mother is too stupid to blame others!

Be all abroad! Should a daughter-in-law be more filial to her mother-in-law than her daughter? There's no need to compare them and ask for trouble!

It is impossible for her to ignore her mother-in-law's attitude, but Bao Bell can say her grievances in front of everyone, not like other men to ask her daughter-in-law to understand her mother-in-law, in fact, more often women care about men's attitudes.

There's really no need to compare my mother-in-law with my mother, I and my mother-in-law are strangers from beginning to end.

This language is ...... organizationalIncoherent and confused.

Solution: (e-t)dt= edt- t dt=et-(1 3)t +c

Find the indefinite integral (e-t) dt

Solution: (e-t) dt=- (e-t) d(e-t)=-(1 3)(e-t) +c

You didn't give. upper and lower bounds, so you can only find indefinite integrals; If you ask for a definite integral, substitute your upper and lower bounds to get it;

Also, your topic is not clear, and it is given to you according to both understandings. 】

That's right, if you take apart the first term in your final result, there will be a -ln3, put -ln3 in the constant c and the answer will be the same.

Ostrovsky's theorem, the integral of a rational function, is divided into an integral part and a pure fractional part, the pure fractional part is divided into a part, the integer part can be obtained by the division of the integer, the method of the partial formula is obtained by the method of the undetermined coefficient or long division, the integral of the integral part is still an integer, after the partial formula, the numerator of all fractions is constant, and the denominator of all fractions is up to a quadratic polynomial (<0 in the case of a quadratic polynomial) The denominator is the integral logarithmic function of a primary polynomial, and the denominator is a quadratic polynomial ( <0) is an arctangent function, so the integral of a rational function must be solvable, and it is composed of an integer, a logarithmic function, and an arctangent function

Let t=arctan x, then x=tan 2t, dx=2tantsec 2tdt

Original = tan 2t*t*2tantsec 2tdt

2∫t*tan^3tsec^2tdt

1/2)*∫t*d(tan^4t)

t/2)*tan^4t-(1/2)*∫tan^4tdt

t/2)*tan^4t-(1/2)*∫tan^2t*(sec^2t-1)dt

t/2)*tan^4t-(1/2)*∫tan^2tsec^2tdt+(1/2)*∫tan^2tdt

t/2)*tan^4t-(1/2)*∫tan^2d(tant)+(1/2)*∫sec^2t-1)dt

t/2)*tan^4t-(1/6)*tan^3t+(1/2)*tant-t/2+c

t/2)*(tan^4t-1)+(1/6)*tant*(3-tan^2t)+c

1 2)*arctan x*(x 2-1)+(1 6)* x*(3-x)+c, where c is an arbitrary constant.

1. Let t= (e x-1), then x=ln(1+t 2), dx=2t (1+t 2)dt

Original = ln(1+t 2)*[1+t 2) t]*[2t (1+t 2)]dt

2∫ln(1+t^2)dt

2tln(1+t^2)-2∫td[ln(1+t^2)]

2tln(1+t^2)-4∫(t^2)/(1+t^2)dt

2tln(1+t^2)-4∫[1-1/(1+t^2)]dt

2tln(1+t^2)-4t+4arctant+c

2x√(e^x-1)-4√(e^x-1)+4arctan√(e^x-1)+c

2(x-2) (e x-1)+4arctan (e x-1)+c, where c is an arbitrary constant.

2. Let t=x, then x=t 2, dx=2tdt

Original = arctant [t(1+t 2)]*2tdt

2∫arctant/(1+t^2)dt

2∫arctantd(arctant)

arctant)^2+c

arctan x) 2+c, where c is an arbitrary constant.

3. Original formula = (1 2) * 1 (x 2-x 2+1) dx

1/√2)*∫1/√[(x-1/4)^2+15/16]d(x-1/4)

1 2)*ln[x-1 4+ (x 2-x 2+1)]+c, where c is an arbitrary constant.

For indefinite integrals, it is normal for the algorithms to be different and the results to be different, but the resulting original function must differ by only one constant. The reason for this is that the result of an indefinite integral is not a number, but a family of functions, and the functions in this family of functions are written as f(x)+c, f(x)+a+c (a is a specific number) is all possible, and c can "absorb" any other real number a.

1. The commutation method, that is, the substitution of bai variables

substitution, and inegral by parts, both of these methods are applicable.

DAO is definite integral, and it is also applicable to indefinite integral. .2. There are many methods that cannot be applied to indefinite integrals, but are applicable to definite integrals.

For example, calculating integrals using retention numbers can only be applied to definite integrals;For the integration of the normally distributed function, the tolerance must be calculated using the generalized integral, that is, the definite integral, in polar coordinates. .3. For indefinite integrals and definite integrals, the third common method is partial fraction. .