What is the original function of 1 X squared under the root number ?

Find the integral for (1+x 2).

Make a triangular substitution, so that x=tant

then (1+x)dx

secttant+ln sect+tant --sect) 3dt, so (sect) 3dx=1 2(secttant+ln sect+tant) +c

Thus (1 x 2) dx

1/2（x√（1+x²）+ln（x+√（1+x²））c<>

Let x=tan(t),t (-pi 2,pi 2), then the root number (1+x 2) = sec(t), and the root number (1+x 2) dx

sec(t)d(tan(t)))- let this integral be i).

tan(t)sec(t)-∫tan(t)d(sec(t))

tan(t)sec(t)-∫tan(t)^

tan(t)sec(t)-∫sec(t)[sec(t)^2-1]dt

tan(t)sec(t)-∫sec(t)d(tan(t))+sec(t)dt

tan(t)sec(t)-∫sec(t)d(tan(t))+ln[sec(t)+tan(t)]

tan(t)sec(t)+ln[sec(t)+tan(t)]-i

So 2i = tan(t)sec(t)+ln[sec(t)+tan(t)]+c

i=/2+c

The 2+c indefinite integral i is the original function that is sought.

The original function of (1+x) is 2 3*(1+x) (3 2)+c. The specific solution process is as follows.

Solution: Let f(x) = (1+x) and f(x) be the original function of f(x).

Then f(x) = (1+x)dx

√(1+x)d(1+x)

2/3*(1+x)^(3/2)+c

That is, the original function of f(x) = (1+x) is f(x)=2 3*(1+x) (3 2)+c.

Geometric meaning. Functions are related to inequalities and equations (elementary functions). Let the value of the function be equal to zero, and from a geometric point of view, the value of the corresponding independent variable is the abscissa of the intersection point of the image and the x-axis; From an algebraic point of view, the corresponding independent variable is the solution of the equation. In addition, if you replace the "=" in the expression of the function (except for functions without expression) with "<" or ">" and replace the "y" with another algebraic formula, the function becomes an inequality and the range of independent variables can be found.

Find the integral for (1+x 2).

Make a triangular substitution, so that x=tant

then (1+x)dx

secttant+ln│sect+tant│--sect)^3dt

So (sect) 3dx=1 2(secttant+ln sect+tant )+c

Thus (1 x 2) dx

Extended information: The original function of the original function refers to the function f(x) of a known function defined in a certain interval, if there is a derivative function f(x), such that df(x)=f(x)dx exists at any point in the interval, then the function f(x) is said to be the original function of the function f(x) in that interval.

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The original function of 1+x 3) is 1 2*(1+x 3) (2)+c. The specific solution process is as follows.

Solution: Let f(x) = (1+x 3) and f(x) be the original function of f(x).

Then f(x) = (1+x 3)dx

√(1+x＾3)d(1+x＾3)

1/2*(1+x＾3)^(2)+c

Question = (1+x 3)d(1+x 3).

How did this step come out? Wrong.

If you look at it as a whole, you might be able to understand it.

You change the next x to x 3 and then 1 has no effect on the function, so it can be 1+x 3

Question: The x behind it can be turned directly into the cubic of x?

So (1 (x 2-x+1))dx= (dx ((x-1 2) 2+(root number 3 2) 2))) = (2 root number 3)arctan((x-1 2) (root number 3 2))+c

This is the complete correct answer, the content is more complicated, you can ask me if you don't understand.

The calculation process is as follows:∫[x/√(1-x²)]dx

½∫1/√(1-x²)]d(1-x²)=-√(1-x²) c

The original function of x(1-x) is - (1-x) cThe existence theorem of the original functionIf the function f(x) is continuous in an interval, then f(x) must exist in that interval, which is a sufficient but not necessary condition, also known as the "original function existence theorem".

For example, x3 is a primitive function of 3x2, and it is easy to know that x3+1 and x3+2 are also primitive functions of 3x2. Therefore, if a function has a primitive function, there are many primitive functions, and the concept of primitive functions was proposed to solve the inverse operation of derivation and differentiation.

For example, if it is known that the velocity of an object moving in a straight line at any time t is v=v(t), the law of its motion is required to find the original function of v=v(t). The problem of the existence of the original function is a fundamental theoretical problem in calculus, and when f(x) is a continuous function, its original function must exist.

The calculation process is as follows:

x/√(1-x²)]dx

½∫1/√(1-x²)]d(1-x²)=-√(1-x²) c

The original function of x(1-x) is - (1-x) c

Just go for points.

The original function is: 1 2 times x times the square of 1-x under the root sign + 1 2 times arcsinx + c (c is an arbitrary constant).

Let x=tan , -2< <2

i.e. dx=sec 2*d

then (1 1+x 2)dx

1／√(1+tanθ^2)*secθ^2*dθ

1/cosθ)dθ

cosθ/(cosθ）^2]dθ

1/[1-(sinθ)^2]d（sinθ)

1/2*ln[（1-sinθ）/1+sinθ)]c

ln[x+ (1+x 2)]+c (c is constant).

Find the original function of 1 root number (1+x 2).

It is to find the integral of the function 1 root number back to the rough skin (1+x 2) to x.

Find the original function of 1 root number (1+x 2) and eliminate the root number (1+x 2) with "trigonometric substitution".

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 the leakage 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 in the stool

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

9、∫ tanx dx = ln|cosx| +c = ln|secx| +c

10、∫ secx dx =ln|cot(x/2)| c

1/2)ln|(1 + sinx)/(1 - sinx)| c

ln|secx - tanx| +c

ln|secx + tanx| +c

Let x=tan(t),t (-pi 2,pi 2), then Gen Pei Chun (1+x 2)=sec(t), and carry the root number (1+x 2)dx= sec(t)d(tan(t))- let this integral be bright in volt i) = tan(t)sec(t)- tan(t)d(sec(t))=tan(t)sec(t)- tan(t)

From the meaning of the title can be obtained:

Potato (1 x) dx = x (-1 2) dx = 2 x + c (c is constant).

So the original function of x under the root number of 1 is 2 Nawang source x+c (c is a constant).