∫1/（X+1)（X+3） dx.

∫1/（X+1)（X+3） dx.

∫ dx/[(x + 1)(x + 3)]
= (1/2)∫ [(x + 3) - (x + 1)]/[(x + 1)(x + 3)] dx
= (1/2)∫ [1/(x + 1) - 1/(x + 3)] dx
= (1/2)[ln| x + 1 | - ln| x + 3 |] + C
= (1/2)ln| (x + 1)/(x + 3) | + C
= ln√[(x + 1)/(x + 3)] + C

Calculation: definite integral ∫ (upper √ 2, lower 0) (√ 2-x ^ 2)

Let x = √ 2 · Sint, then DX = √ 2 · costdt
∫（0→√2）√(2-x²)dx
=∫（0→π/2）√2·cost·√2·cost dt
=∫（0→π/2）2·cos²t dt
=∫（0→π/2）(1+cos2t) dt
=[t+1/2·sin2t]|（0→π/2）
=π/2

Find: definite integral ∫ (1,0) e ^ - x ^ 2 DX

The indefinite integral of e ^ (- x ^ 2) can not be expressed by elementary function. There are two methods: one is to use numerical integration method, such as Simpson method, for the definite integral of ∫ (0 → 1) e ^ (- x ^ 2) DX; the other is to expand e ^ (- x ^ 2) series, integrate term by term, and then calculate the value of the definite integral ∫ e ^ (- x ^ 2) DX = ∑ (n: 0 →∞) (- 1) ^ n * x ^ (2n + 1) / [...]

Calculation: definite integral ∫ (upper 1, lower 0) x / 1 + x ^ 2

∫(0→1) x/(1 + x²) dx
= (1/2)∫(0→1) d(1 + x²)/(1 + x²)
= (1/2)ln(1 + x²) |(0→1)
= (1/2)ln(1 + 1)
= (1/2)ln(2)

Calculation: definite integral ∫ (upper √ 2, lower 0) (√ 2-x ^ 2)

Understanding the practical significance of definite integral
This question y = √ (2-x & # 178;)
x²+y²=2(y≥0)
Represents the upper half of a circle
So the value of definite integral is 1 / 4 area of circle
So it is (1 / 4) * π * (√ 2) &# 178; = π / 2

Calculation: definite integral ∫ (up-2, Down-3) 1 / 1 + X ∫

∫ (up-2, Down-3) 1 / 1 + X DX
=ln|1+x|[-3,-2]
=-ln2

Calculation: definite integral ∫ (in the upper 9, in the lower 4) √ x (1 + √ x) DX to find the detailed process answer

∫（4→9）√x(1+√x)dx
=∫（4→9）(√x+x) dx
=[2/3·x^(3/2)+x²/2] |（4→9）
=[2/3·9^(3/2)+9²/2]-[2/3·4^(3/2)+4²/2]
=271/6

∫ (3 sin T + 1 / 2 sin ^ 2 T) DT

The original formula = 3 ∫ sintdt + ∫ CSC & # 178; TDT
=-3cost-cott+C

∫ t^2 * sin(t) dt

∫ t^2 * sin(t) dt
=- ∫ t²dcost
=-∫ t²cost+ ∫costd t²
=- t²cost+ 2∫tcostdt
=- t²cost+ 2∫tdsint
=- t²cost+ 2tsint-2∫sintdt
=- t²cost+ 2tsint+2cost+C

Why does g (x) = x + 2 + sin (x + 1) take (- 1,1) as the center of symmetry

f(x)=x+sinx (x∈R)
Then f (- x) = - x + sin (- x) = - x-sinx = - f (x)
So f (x) is an odd function, symmetric with respect to (0,0),
How to change the f (x) image to h (x) = f (x + 1) = x + 1 + sin (x + 1) by moving it one unit to the left along the X axis
If the H (x) image is moved up one unit along the Y axis, it becomes g (x) = x + 1 + sin (x + 1) + 1 = x + 2 + sin (x + 1)
Then the center of symmetry moves one unit to the left along the x-axis to (- 1,0), and then moves one unit up along the y-axis to (- 1,1)