1 x under the root number 1 x dx 2 x 2e x dx 3 xarctanx dx 4 xsin2x dx thanks! process

1 ream (1 x) u, get: x u 2 1, dx 2udu.

Original u 2 1) u (2u)du

2∫（u^2－1）du

2∫u^2du－2∫du

2/3）u^3－2u＋c

2/3）（1＋x）√（1＋x）－2√（1＋x）＋c

x^2e^x－2xe^x＋2e^x＋c

3 Original (1 2) Arctanxdx 2 (1 2) x 2Arctanx (1 2) x 2D (Arctanx).

1/2）x^2arctanx－（1/2）∫［x^2/（1＋x^2）］dx

1/2）x^2arctanx－（1/2）∫dx＋（1/2）∫［1/（1＋x^2）］dx

1/2）x^2arctanx－x/2＋（1/2）arctanx＋c

4. Original (1 2) xsin2xd(2x) 1 2) xd(cos2x) 1 2)xcos2x (1 2) cos2xdx

1/2）xcos2x＋（1/4）∫cos2xd（2x）＝－1/2）xcos2x＋（1/4）sin2x＋c .

Question 1: Let (1 x) u, get: x u 2 1, dx 2udu.

Original u 2 1) u (2u)du

2∫（u^2－1）du

2∫u^2du－2∫du

2/3）u^3－2u＋c

2/3）（1＋x）√（1＋x）－2√（1＋x）＋c

Question 2: Original x 2de x x 2e x e xdx 2 x 2e x 2 xe xdx 2e x 2 xde x 2e x 2e x 2 e xdx

x^2e^x－2xe^x＋2e^x＋c

Question 3: Original (1 2) arctanxdx 2 (1 2) x 2arctanx (1 2) x 2d(arctanx).

1/2）x^2arctanx－（1/2）∫［x^2/（1＋x^2）］dx

1/2）x^2arctanx－（1/2）∫dx＋（1/2）∫［1/（1＋x^2）］dx

1/2）x^2arctanx－x/2＋（1/2）arctanx＋c

Question 4: Original (1 2) xsin2xd(2x) 1 2) xd(cos2x) 1 2)xcos2x (1 2) cos2xdx

1/2）xcos2x＋（1/4）∫cos2xd（2x）＝－1/2）xcos2x＋（1/4）sin2x＋c

Summary. For this differential equation, it can be solved using the solution method of calculus. First, the derivative of the equation can be obtained:

d dx (d dx (x 2sin(x+1)))d 2 dx 2 (x 2sin(x+1)) d2x 2sin(x+1)dxTherefore, the differential equation can be rewritten as: d 2 dx 2 (x 2sin(x+1)) d2x 2sin(x+1)dxNext, we need to solve the equation. First, find the first derivative of x 2sin(x+1), and we get:

d dx (x 2sin(x+1)) 2xsin(x+1) +x 2cos(x+1) and then the derivative of the first derivative is obtained as follows: d 2 dx 2 (x 2sin(x+1)) 2sin(x+1) +4xcos(x+1) +x 2(-sin(x+1)) 2xcos(x+1).

9. d2x^2sin(x+1)dx=

Question 9. For this differential equation, the method of calculus can be used to solve the problem. The hand volt first derives the equation, which can be obtained:

d dx (d dx (x 2sin(x+1)))d 2 dx 2 (x 2sin(x+1)) d2x 2sin(x+1)dxTherefore, the differential equation can be rewritten as: d 2 dx 2 (x 2sin(x+1)) d2x 2sin(x+1)dxNext, we need to solve the equation. First, find a derivative of x 2sin(x+1) with potatoes, and we get:

d dx (x 2sin(x+1)) 2xsin(x+1) +x 2cos(x+1) and then the derivative of the first derivative is obtained as follows: d 2 dx 2 (x 2sin(x+1)) 2sin(x+1) +4xcos(x+1) +x 2(-sin(x+1)) 2xcos(x+1).

Substituting the above results into the original differential equation, we get: (2sin(x+1) +4xcos(x+1) +x 2(-sin(x+1)) 2xcos(x+1)) d2x 2sin(x+1)dxTherefore, the solution of the differential equation is: x 2sin(x+1) =d2x 2sin(x+1)dx)dx + c1]dx + c2 where c1 and c2 are constants.

It should be noted that for this differential equation, the specific form of the solution may be affected by the condition of being ready to the beginning of the first skin, so it is necessary to determine the specific form of the solution according to the initial condition in the practical problem.

9. d[2x^2sin(x+1)dx=

Hello, for this problem, we can use the derivative rule to solve it. According to the product rule, we can split d[2x 2sin(x+1)dx] into 2x 2d[sin(x+1)dx]+sin(x+1)d[2x 2dx]. Then according to the chain rule, we can find d[sin(x+1)dx]=cos(x+1)dx and d[2x 2dx]=4xdx.

Therefore, the original formula can be reduced to 2x 2cos(x+1)dx+4x 2sin(x+1)dx. As we can see from the title, this is a mathematical problem, and mathematics is a very important subject, it is not only a tool, but also a way of thinking. By learning mathematics, we can exercise our logical thinking and analytical skills, and improve our problem-solving skills.

At the same time, mathematics is also the foundation of many other disciplines, such as physics, chemistry, computer science, and many more. In addition to mathematics, we can also see an important concept in this problem - derivatives. Derivatives are an important concept in calculus that can be used to describe the rate of change of a function over a certain point of age.

Derivatives have a wide range of applications in many fields, such as physics, economics, engineering, and more. In conclusion, mathematics and derivatives are very important concepts, and they are not only important in the academic field, but also an integral part of our daily lives. Hopefully, mine will help you better understand this question and related concepts.

Summary. Classmate, I'm Mr. Mumu, and now I can send me the question.

9. d[2x^2sin(x+1)dx=

Classmate, I'm Mr. Mumu, and now I can send me the question.

Hurry up, I'm in a hurry.

Good. Question 9.

What do you mean, both of these are the first answers?

Well, the ** just now, the teacher will give you the method of finding the derivation of the composite function, so that you can easily understand.

Student, there is a classmate just now who is exactly the same as the three questions you asked<>

9.+d[2x^2sin(x+1)dx=

Hello, +d[2x 2sin(x+1)dx= (1 2)sin(x 2+1)d(x 2) =1 2) sin(x 2+1)d(x 2+1) is also the difference is to put the x in front of sin into d to become d(x 2), then the outside should be multiplied by 1 2 to be the same as the original question, and d(x 2) and d(x 2+1) are the same, if x 2+1 is made to t, it becomes to find (1 2) sin(t)d(t) The integration of the disadvantages of slippery, this is to make up the differential method of renting Qingla.

Changing the Yuan Law to Honor the Wheel.

Let (x+1)=t

x=t^2-1

dx=2tdt

1/√(x+1)dx

1/t*2tdt

Group shirt 2dt2t+c2 collapsed cavity (x+1)+c