How to find the indefinite integral of ∫ (x ^ 2) DX / (x + 1) (x + 3)? Well, I'm a bit of a nerd, so I'm not very good at it It is better to use the method of ∫ u'v = UV - ∫ UV '

How to find the indefinite integral of ∫ (x ^ 2) DX / (x + 1) (x + 3)? Well, I'm a bit of a nerd, so I'm not very good at it It is better to use the method of ∫ u'v = UV - ∫ UV '

∫1/(X+1）（x+3） dx
=∫dx/2(x+1)-∫dx/2(x+3)
=1/2*∫d(x+1)/(x+1)-1/2*∫d(x+3)/(x+3)
=ln(x+1)/2-ln(x+3)/2+C
=1/2*[ln(x+1)-ln(x+3)]+C
=Ln [(x + 1) / (x + 3)] / 2 + C note that denominator 2 is not in LN, but out of LN
It is known that f (x) is a decreasing function on the real number set R, and f (x / y) = f (x) - f (y), f (2) = 1. The solution to the inequality is f (x) + F (X-7) > = 3
First of all, let x = y = 1, f (1) = 0; f (2) = 1, so f (1 / 2) = f (1) - f (2) = - 1, f (2) - f (1 / 2) = f (4) = 2, so 3 = f (2) + F (4), so f (x) + F (X-7) > = 3, that is: F (x) + F (X-7) > = f (2) + F (4), shift term, f (x) - f (2) > = f (4) - f (X-7), so f (x / 2) > = f (4 / (X-7)) because f (x) is R
f(x/y)=f(x)-f(y),f(2)=1,
f(2/1)=f(2)-f(1),f(1)=0
f（4）=2,f(8)=3
f(x)>=3-f(x-7), f(x)>=f(8)-f(x-7),
f(x)>=f(8/(x-7)),
X
Find the definite integral of | cosx | DX from the lower limit 0 to the upper limit
The original formula = ∫ (0, π / 2) cosxdx - ∫ (π / 2, π) cosxdx
=(sinx)│(0,π/2)-(sinx)│(π/2,π)
=(1-0)-(0-1)
=2
Given the function f (x) = ① (x-1) &# 178; + 2, X ≤ 1, ② 2, x > 1, find the solution set of the inequality f (1-x & # 178;) > F (2x)
Finding definite integral ∫ [0,2 π] √ (1 + cosx) DX
1+cosx=1+(2cos²(x/2)--1]=2cos²(x/2)
Zero
2 π ask: the process.
The known function f (x) = x & # 178; + 2 / X (x ≠ 0) solves the inequality f (x) - f (x-1) > 2x-1
f(x-1)=（x-1）² + 2/（x-1）
f(x)-f(x-1)=x²+2/x - （x-1）² - 2/（x-1）=2x - 1 + 2/x - 2/（x-1）
If f (x) - f (x-1) > 2x-1 is reduced to 2x-1 + 2 / X-2 / (x-1) > 2x-1
2/x - 2/（x-1）＞0
x（x-1）
Two intersections ab of the image of quadratic function y = x & # 178; + (M + 1) x + m and X axis, and ab = 2, what is m = then
When y = 0, X1 = - 1, X2 = - M
∴AB=|-1+m|=2
∴m1=3 m2=-1
We know that the analytic expression of function f (x) is 2x & # 178; + X, and solve the inequality [f (x)] &# 178; + 2x & # 178; + X
[f(x)]²+2x+x＜0
[f(x)]²+f(x)＜0
f(x)(f(x)+1)＜0
-1＜f(x)＜0
-1＜2x²+x＜0
2x²+x+1＞0①
x(2x+1)＜0②
①x∈R②-1/2＜x＜0
The solution set {x | - 1 / 2 ＜ x ＜ 0} is obtained from (1) and (2)
Given the quadratic function y = 2x square - x + m, (1) what is the value of M, the vertex is above the x-axis. (2) if the throwing line intersects with the axis a through a, make another point B of the AB / / axis throwing line, and when s △ AOB = 4, find the analytic expression of the quadratic function
(1) When B ^ 2-4ac0, B0, the analytical formula of parabola is
y=x^2-x+16
When m
Given the function f (x) = SiNx + 2x, X ∈ R, if f (1-A) + F (2a)
Because f (x) is an odd function, f (1-A) + F (2a)
So hard!