∫(x^3-3x^2+4x-9)/(x^2+3)dx

∫(x^3-3x^2+4x-9)/(x^2+3)dx

(x^3-3x^2+4x-9)/(x^2+3)=x-3+x/(x^2+3)
The original formula = ∫ xdx - ∫ 3DX + ∫ X / (x ^ 2 + 3) DX
=x^2/2-3x+1/2∫d(x^2+3)/(x^2+3)
=x^2/2-3x+ln(x^2+3)/2+C

(3x ^ 2 + 1) / ((x ^ - 1) ^ 3) DX integral
Ask for all kinds of methods. Don't be stingy!
The denominator is x ^ 2

Pro, it's not hard. Look
Let t = X-1
Then the original formula = [3 (T + 1) ² + 1] / T & #179;
Well organized
=3/t + 6/t² + 4/t³
Integral gain
3ln |t| - 6/t - 2/t² +C
= 3ln |x-1| - 6/(x-1) - 2/(x-1)² +C

Finding integral ∫ (3x + 1 / x∧2) DX
It's all wrong

∫（3x+1/x∧2)dx=-∫（3x+1)d(1/x)
=-(3x+1)/x+∫（1/x)d(3x+1)
=-(3x+1)/x+1/3∫（1/x)dx
=-(3x+1)/x+1/3ln|x|+c
That's it!

In mathematics, when the following formula of quadratic function is greater than, less than or equal to zero, how is the root distributed

It's not a root formula, it's a discriminant
△=b^2-4ac
When it is greater than 0, it has two unequal real roots; when it is equal to 0, it has two equal real roots; when it is less than 0, it has no real roots, so it has no solution in junior high school; when it studies imaginary numbers and complex numbers in senior high school, it has a pair of conjugate complex roots

The height of the cone is 10cm, the side view is semicircle, and the side area of the cone is 5cm______ ．

Let the generatrix length of the cone be l, ∵ the expanded side view be a semicircle, ∵ π L = 2 π R, ∵ r = L2, ∵ the height of the cone be 10cm, ∵ (10) 2 + R2 = L2, the solution is: r = 1033cm, ∵ generatrix L = 2033cm, ∵ the side area of the cone is: 2 π × 1033 × 2033 △ 2 = 2003 π cm2

In division with remainder, the divisor is the quotient product______ Plus______ ．

In the divisor formula with remainder, the formula is: divisor = quotient × divisor + remainder, so the divisor is equal to quotient, multiplier, divisor and remainder. So the answer is: divisor, quotient

The circumference of a circle is 12.56 cm, and its radius is______ Cm, the area is______ Square centimeter

12.56 △ 3.14 △ 2 = 2 (CM); 3.14 × 22 = 3.14 × 4 = 12.56 (square cm); answer: the radius of this circle is 2 cm and the area is 12.56 square cm

As shown in the figure, it is known that s is a point out of the plane of the parallelogram ABCD, m and N are points on SA and BD respectively, and SMMA = bnnd. Then the line Mn______ Plane SBC

It is proved that bnnd = bgag can be obtained by making ng ‖ ad through N, intersecting AB with G and connecting mg. According to the known condition bnnd = SMMA, SMMA = bgag, ⊄ mg ‖ sb. ⊄ mg ⊄ plane SBC, sb ⊂ plane SBC, ⊂ mg ‖ plane SBC. Ad ‖ BC, ⊄ ng ‖ BC, ng ⊄ plane SBC, BC ⊂ plane SBC ⊂ ng ‖ plane SBC

The perimeter of the sector of radius 4cm is equal to the perimeter of the semicircle where the arc is located

Let the arc length be l and the radius be r = 4
Then l + 8 = 4
So l = 4-8
So the sector area s = (1 / 2) RL = (1 / 2) * 4 * (4 pai-8) = 8 pai-16

In the arithmetic sequence {an}, A1 = 2, tolerance D ≠ 0, and A1, A3 and a11 are just the first three terms of an arithmetic sequence, then the common ratio of the arithmetic sequence is ()
A. 2B. 12C. 14D. 4

In the arithmetic sequence {an}, A1 = 2, A3 = 2 + 2D, a11 = 2 + 10d. Because A1, A3 and a11 are just the first three terms of an arithmetic sequence, there is A32 = a1a1a11, that is, (2 + 2D) 2 = 2 (2 + 10d), and the solution is d = 3, so the common ratio of the arithmetic sequence is 82 = 4, so D is selected