The integral of ∫ 3x + 1 / x ^ 2 + 4x + 5 DX

The integral of ∫ 3x + 1 / x ^ 2 + 4x + 5 DX

∫ (3x+1)dx/(x^2+4x+5) = ∫ (3x+6-5)dx/(x^2+4x+5) = (3/2)∫ (2x+4)dx/(x^2+4x+5) - 5 ∫ dx/(x^2+4x+5)= (3/2))∫ d(x^2+4x+5)/(x^2+4x+5) -5 ∫ d(x+2)/[1+(x+2)^2]= (3/2)ln(x^2+4x+5) - 5arctan(x+2)+C

∫(1-x)/[√(9-4x^2)]dx=

∫ (1-x) / [√ (9-4X ^ 2)] DX = 5 - there are still 14 days and 23 hours to the end of the question, kkcl111 - trial period first level answer substitution order x = 3 / 2sint, t ∈ [- 0.5 π, 0.5 π] to get ∫ (1-x) / [√ (9-4X ^ 2)] DX = ∫ (1-1.5sint) 1.5costdt/3cost = ∫ (1-1.5sint

∫1-x/(√9-4x^2)dx,

 
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If ray AB and ray CD intersect at point P, is APC called an angle?
The problem is that point a is the vertex of ray AB, point C is the vertex of ray CD, and P is the intersection point. A and C are fixed and cannot be extended. Then it is obvious that AP and CP are just line segments and cannot be regarded as ray,

The eighth book of PEP describes the angle as follows: the figure composed of two rays drawn from one point is called angle
Although AP and CP are line segments, they can be thought of as two points intercepted on two rays. Two points intercepted on a ray do not affect the size of the angle, so obviously APC is an angle
Of course, the PEP has some problems with the definition of diagonal. I have no personal opinion
It can't be a ray, but it can be imagined as two points intercepted on the ray. Imagine. Imagine. Don't fall asleep

In the rectangular coordinate system, we know a (- 4,6), B (2,0), C (1,0) d (- 2,3) to find the value of AB: CD

∵AB＝√[(-4-2)²+(6-0)²]=6√2
CD=√[(-2-1)²+(3-0)²]=3√2
∴AB：CD＝(6√2)：(3√2)
　　　　　＝2

Given that a (- 3, - 2), B (2, - 2), C (- 1,1), D (2,1) are four in the coordinate plane, then the position relationship between the line AB and the line CD is?
Come on

Look, the ordinate of AB is the same. So ab ‖ X axis
The ordinate of CD is the same. So CD ‖ X axis
So ab ‖ CD

If ab ⊥ CD, AC ⊥ BD and ad ⊥ BC are satisfied at the same time, then the position relationship of a, B, C and D is

If you draw a cube, take any point on the top side as a, and take three points on the opposite side to form a regular pyramid

It is known that a.b.c.d is the four vertexes of a quadrilateral in space

If AB and CD are parallel or intersect, then AB and CD are coplanar, and the four points of ABCD are in the same plane. This is contrary to the proposition, so the hypothesis is not tenable and the proposition is correct

If the points a (1, - 2, - 3), B (- 1, - 1, - 1) and C (0,0, - 5) are known in the space rectangular coordinate system, then the shape of △ ABC is

ab=3
ac=3
bc=3√2
isosceles right triangle

In the space rectangular coordinate system, given the fixed point m (2, - 1, 3), if point a and point m are symmetrical about the xoy plane, and point B and point m are symmetrical about the X axis, then | ab | = ()
A. 2B. 4C. 25D. 37

∵ point m (2, - 1,3) about plane xoy symmetry point a, its abscissa and ordinate are invariable, vertical coordinate is opposite, so a (2, - 1, - 3); the symmetry points of M (2, - 1,3) about X axis are respectively B, its abscissa is invariable, ordinate is opposite, vertical coordinate is opposite, B (2,1, - 3), | ab | = (2 − 2) 2 + 22 + (− 3 + 3) 2 = 2, so select a