Finding definite integral ∫ ln [x + √ (X & # 178; + 1)] DX x belongs to [0,2]

Finding definite integral ∫ ln [x + √ (X & # 178; + 1)] DX x belongs to [0,2]

The answer is 2ln (2 + √ 5) - √ 5 + 1
∫(0~2) ln[x + √(x² + 1)] dx
= { xln[x + √(x² + 1)] } |(0~2) - ∫(0~2) x dln[x + √(x² + 1)]
= 2ln(2 + √5) - ∫(0~2) x • 1/[x + √(x² + 1)] • [1 + x/√(x² + 1)] dx
= 2ln(2 + √5) - ∫(0~2) x/√(x² + 1) dx
= 2ln(2 + √5) - (1/2)∫(0~2) 1/√(x² + 1) d(x² + 1)
= 2ln(2 + √5) - (1/2) • 2√(x² + 1) |(0~2)
= 2ln(2 + √5) - √5 + 1
∫ln[x+√(x²+1）] dx
=xln[x+√(x²+1）]|[0,2]- ∫[0,2]xdln[x+√(x²+1）]
=xln[x+√(x²+1）]|[0,2]- ∫[0,2]x/√(x²+1）dx
=2ln(2+√5)-2(x²+1)^(1/2)|(0,2)
=2ln (2 + √ 5) - 2 √ 5 + 2DL... Expansion
∫ln[x+√(x²+1）] dx
=xln[x+√(x²+1）]|[0,2]- ∫[0,2]xdln[x+√(x²+1）]
=xln[x+√(x²+1）]|[0,2]- ∫[0,2]x/√(x²+1）dx
=2ln(2+√5)-2(x²+1)^(1/2)|(0,2)
=How does DLN [x + √ (X & # 178; + 1)] equal to √ (X & # 178; + 1) DX? In more detail, the answer is 2ln (2 + √ 5) - √ 5 + 1
Try to compare the average change rate of sine function y = SiNx in the interval [0, π / 6] and [π / 3, π / 2], and compare the size
The average rate of change of Δ Y1 = f (π / 6) - f (0) = 1 / 2 in the interval [0, π / 6] V1 = Δ Y1 / Δ X1 = (1 / 2) / (π / 6) = 3 / π Δ y2 = f (π / 2) - f (π / 3) = 1 - √ 3 / 2 in the interval [π / 3, π / 2] V2 = Δ Y2 / Δ x2 = (1 - √ 3 / 2) / (π / 6) = 6 / π - 3 √ 3 / π V1-V2 = 6 / π + 3 √ 3 / π >
Calculate definite integral ∫ (0,1) x ^ 2 · √ (4-3x ^ 3) · DX
∫(0、1)x^2·√(4-3x^3)·dx
=-1/9 ∫(0、1)√(4-3x^3)·d(4-3x^3)·
=[- 1 / 9 * 2 / 3 (4-3x ^ 3) ^ (3 / 2)] (lower limit 0, upper limit 1)
=14/27
Why can't you see the sign clearly.. According to the answer from upstairs.. I'll do the math. The answer is 14 / 27
Y = SiNx, SiNx > 0 find the interval of x.sinx
Draw the unit circle, then the value of Y corresponding to the point on the unit circle is SiNx, and the value of X is cosx
When the point is in one or two quadrants, SiNx ＞ 0 (because it represents y, y is positive in one or two quadrants of course), so the interval of SiNx ＞ 0 is (2k Π, Π + 2K Π)
The interval less than 0 is (Π + 2K Π, 2 Π + 2K Π)
Or you can draw the function image of SiNx, which is more intuitive
The range of [2K π, 2K π + π] > 0 is the complementary interval
How to calculate the ∫ 1 / (XLN ^ 3x) DX definite integral
How to calculate ∫ 1 / (XLN ^ 3x) DX indefinite integral
∫1/(xln^3x) dx
=∫1/(lnx)^3 d(lnx)
=-(1/2)∫d(lnx)^(-2)
=-1/(2(lnx)^2) +C
If you want to, you can make progress only if you want to
When 0 ＜ x ＜ Pai / 2, try to find SiNx ＞ x-x & # / 6
It should be a proof
prove:
Constructor f (x) = SiNx - (x-x & #³ / 6)
∴ f(0)=0
f'(x)=cosx-(1-x²/2)=g(x)
∴ g(0)=0
Then G '(x) = - SiNx + X
∵ 00
∴ 00
∴ sinx＞x-x³/6
For definite integral ∫ upper limit 1, lower limit 0 (3x ^ 4 + 3x ^ 2 + 1) / (x ^ 2 + 1) DX,
For convenience, the upper and lower limits of the integral are not written, and they are replaced at last
Original formula = ∫ {3x ^ 2 (x ^ 2 + 1) + 1} / (x ^ 2 + 1) DX
=∫3x^2 dx +∫1/(x^2+1)dx
= x^4| +arctanx|
=1 + Pai / 4
Pai is the 3.14
The original function of ∫ 1 / (x ^ 2 + 1) DX is arctanx + C
I hope it can help you
What is the range of the function y = SiNx, X ∈ [- 1 / 3 faction, 6 / 5 faction]?
If you draw a graph, the minimum value is negative root sign 3 / 2 (- 60 degrees), and the maximum value is 1 (90 degrees)
The process of finding definite integral of ∫ lower limit-1 upper limit 3 (3x2-2x + 1) DX
SiNx = 1 / 2, then x = π / 6, x = 5 π / 6 why
SiNx = 1 / 2, then x = π / 6, x = 5 π / 6, because sin π / 6 = 1 / 2.. sin π / 6 = sin (π - π / 6) = sin 5 π / 6 = 1 / 2. In fact, this is only the solution in one period of [0,2 π]
If in the whole range of real numbers, the solution set is: {X / x = π / 6 + 2K π or x = 5 π / 6 + 2K π, K ∈ Z}
SiNx = 1 / 2, x = 30 degrees, π = 180 degrees.