Indefinite integral ∫ 1 / (x ^ 2 + 4) DX, master problem solving steps, thank you

Indefinite integral ∫ 1 / (x ^ 2 + 4) DX, master problem solving steps, thank you

∫1/(x^2+4)dx
=1/4∫1/[1+(x/2)²]dx
=1/2∫1/[1+(x/2)²]d(x/2)
=1/2arctanx/2+c
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∫1/(x^2+4)dx
=1/4∫1/[1+(x/2)²]dx
=1/2∫1/[1+(x/2)²]d(x/2)
=1/2arctanx/2+c
They have talked about the differential and the exchange method
Trigonometric substitution
Let x = 2 Tan U
dx = 2 sec^2 u du
∫1/(x^2+4)dx
= ∫2sec^2 u/(4sec^2 u) du
= (1/2)u + c
= (1/2) tan^-1(x/2) + c
The direct formula of differential
∫1/(x^2+4)dx
=1 / 4 ∫ 1 / [... Expanded
They have talked about the differential and the exchange method
Trigonometric substitution
Let x = 2 Tan U
dx = 2 sec^2 u du
∫1/(x^2+4)dx
= ∫2sec^2 u/(4sec^2 u) du
= (1/2)u + c
= (1/2) tan^-1(x/2) + c
The direct formula of differential
∫1/(x^2+4)dx
=1/4∫1/[1+(x/2)²]dx
=1/2∫1/[1+(x/2)²]d(x/2)
=1 / 2arctanx / 2 + C ﹣ Stow
Factorization - AB (a-b) & # 178; + a (B-A) & # 178; - AC (a-b) & # 178=
First of all, we should be clear about (a-b) &# 178; = (B-A) &# 178;
Therefore, the above formula extracts the common factor A (a-b) ²
We obtain a (a-b) ² * (- B + 1-C)
Detailed steps of solving indefinite integral ∫ x ^ 2 / [√ (2-x)] DX
Let √ (2-x) = t
x=2-t^2 dx=d(2-t^2)=-2tdt
Substituting
∫x^2/[√(2-x)]dx=∫(8t^2-2t^4-8)dt=(8/3)t^3-(2/5)t^5-8t+C
T = √ (2-x) can be substituted
Decomposition factor: (X & # 178; - 2XY) &# 178; + 2Y & # 178; (X & # 178; - 2XY) + y ^ 4
=(x²-2xy+y²)²
=The fourth power of (X-Y)
Correct answer (X-Y) ^ 4
Finding indefinite integral ∫ DX / (1 + x ^ 1 / 2)
Let √ x = t
x=t^2
dx=2tdt
∫dx/(1+x^1/2 )
=∫1/(1+t)*2tdt
=2∫[1-1/(1+t)]dt
=2t-2ln(1+t)+C
Let's go against it
X & # 178; - 2XY + Y & # 178; + 2x-2y + 1 factorization factor
X & # 178; - 2XY + Y & # 178; + 2x-2y + 1 factorization factor
=（x-y）²+2(x-y)+1
=[(x-y)+1]²
=(x-y+1)²;
I'm very glad to answer your questions. Skyhunter 002 will answer your questions
If you don't understand this question, you can ask,
x²－2xy＋y²＋2x－2y＋1
=（x－y）²＋2（x－y）＋1
=（x－y＋1）²
X & # 178; - 2XY + Y & # 178; + 2x-2y + 1 factorization factor
=（x-y）²+2(x-y)+1
=[(x-y)+1]²
=(X-Y + 1) &; en en, I'm glad to help you answer. I wish you all the best and happy every day
Finding definite integral (lower limit 0, upper limit π / 4) ∫ (1 / (1 + (cosx) ^ 2)) DX
Decomposition factor: Y (y + 1) (X & # 178; + 1) + X (2Y & # 178; + 2Y + 1)
y（y+1）（x²+1）+x（2y²+2y+1）
=y(y+1)x²+(2y²+2y+1)x+y(y+1)
=(yx+y+1)[(y+1)x+y]
=(xy+y+1)(xy+x+y)
Finding definite integral [upper limit π, lower limit 0] ∫ (x ^ 2) SGN (cosx) DX
[0, PI / 2]
sgn(cosx) = 1
[pi / 2, PI]
sgn(cosx) = -1
therefore
∫(x^2) sgn(cosx)dx
=∫[0,Pi/2](x^2) dx-∫[Pi/2,Pi](x^2)dx
=Pi^3/24 - Pi^3/3 + Pi^3/24
=-Pi^3/4
∫(0,π)(x^2)sgn(cosx)dx=∫(0,π/2)(x^2)dx-∫(π/2,π)(x^2)dx=-(1/4)π^3
Decomposition factor: x2-y2-2y-1=______ .
x2-y2-2y-1，=x2-（y2+2y+1），=x2-（y+1）2，=（x+y+1）（x-y-1）．