Finding the indefinite integral of 1 / sqrt (x ^ 2 + 1)

Finding the indefinite integral of 1 / sqrt (x ^ 2 + 1)

dx /√(x^2+1)
=∫ [x+√(x^2+1)] /{√(x^2+1)*[x+√(x^2+1)]} dx
=∫ [1 + x/√(x^2+1)]dx /[x+√(x^2+1)]
=∫ d[x + √(x^2+1)] /[x+√(x^2+1)]
= ln[x+√(x^2+1)] + C

1 / 2,1 / 6, (), 9 / 2,27 / 2

1/6,1/2,（3/2）,9/2,27/2
If you multiply each term by 3, you get the next term
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For B of fraction a (a, B is a non-zero natural number), when a (), it is a true fraction, when a (), it is a false fraction,

For B of fraction a (a, B is a non-zero natural number), when a (greater than B), it is a true fraction, and when a (less than or equal to b), it is a false fraction

Are all natural numbers integers
But integers include negative integers

Yes, all natural numbers include 0 and integers
You're talking about natural numbers including integers, not integers including natural numbers, all not negative integers

Find the rule: 1.1.3.5.11.21.43 ()

85,
The number on the odd digit is equal to the double of the previous number plus 1,3 = 2 × 1 + 1,11 = 2 × 5 + 1,43 = 2 × 21 + 1
The number in the even number position is equal to 2 times of the previous number minus 1,1 = 2 × 1-1,5 = 2 × 3-1,21 = 2 × 11-1, so the number in () is in the even number position, so () = 2 × 43-1 = 85,

In the form of 1990, 1990 What is the minimum natural number n that can be divisible by 11?

19901990… 1990 138 (n 1990)
The sum of odd digits is 10N + 9
The sum of even digits is 9N + 3
It is easy to see that 10N + 9 is greater than 9N + 3
So 10N + 9 - (9N + 3) = n + 6
And because
19901990… 1990 138 (n 1990)
Divide by 11
So n + 6 is divided by 11
So n = 5

The image with exponential function y = f (x) = a ^ x passes through point a (1,3)
1) Finding the value of a
2) If y ∈ (1 / 3,9] find the value range of X

1.
a^1=3
a=3
two
y∈（1/3,9]
∴1/3

I know what should I do next--------

I know what to do next
I know what I should do next
I know what I need to do.
I know what I have to do.
I know what I ought to do.
I know what I planned to do.

If the quadratic equation x2 + 4x + k = 0 of X has a real solution, then the value range of K is ()
A. k≥4B. k≤4C. k＞4D. k=4

∵ the quadratic equation of one variable x2 + 4x + k = 0 with respect to X has a real solution, b2-4ac = 42-4 × 1 × K ≥ 0, the solution is k ≤ 4, so B is chosen

That, the application problem of quadratic equation of one variable
There is a question: the output of a farm group will increase by 44% in two years. If the average growth rate is the same in two years, what is the growth rate?

Let the growth rate be X
(1+x)^2=1+44%
1 + x = 1.2 or - 1.2
X = 0.2 or - 2.2
So x = 20%
A, the growth rate is 20%