Solving indefinite integral ∫ 1 / (2x-3) ^ 2DX by substitution method

Let 2x-3 = t, x = (T + 3) / 2, then ∫ 1 / (2x-3) ^ 2DX = ∫ 1 / T ^ 2D [(T + 3) / 2] = ∫ 1 / 2T ^ 2DT = - 1 / 2T + C, that is ∫ 1 / (2x-3) ^ 2DX = - 1 / (4x-6) + C

Solving indefinite integral with the second transformation method: ∫ x ^ 2DX / √ 1-x ^ 2

Let: x = sint
∫x^2dx/√1-x^2
=∫sin^2t costdt /cost
=∫sin^2t dt
=1/2∫(1-cos2t)dt
=t/2-sin2t/4 +c
=t/2-sintcost/2+c
=1/2[arcsinx - x√1-x^2]+c

Solving indefinite integral ∫ [e ^ (1 / x)] / x ^ 2DX by substitution method

Use the substitution integral method:
Method 1:
∫(1/x ²) (e^1/x)dx
Let t = 1 / x, DT = (- 1 / x) ²) dx,dx=(-x ²) DT, substitute DX, about X ²
=∫e^t*(-1)dt
=-∫(e^t)dt
=-e^t+C
=-(e^1/x)+C
Method 2:
∫(1/x ²) (e^1/x)dx
d(1/x)=(-1/x ²) dx,∴dx=(-x ²) D (1 / x), substitute DX
=∫(1/x ²) (e^1/x)*(-x ²) d(1/x)
=-∫(e^1/x)d(1/x)
=-(e^1/x)+C

Indefinite integral: ∫ x ^ 2DX / root sign (a ^ 2-x ^ 2)=

Let x = asint, then DX = acost DT ∫ x ²/ √(a ²- x ²) dx=∫a ² sin ² t/(acost)·acost dt=a ² ∫sin ² t dt=a ² ∫(1-cos2x)/2 dt=a ² [t-1/4·sin2x]+C=a ² [arcsin(x/a)-1/2·x/a·√(1-...

√ a ^ 2 + x ^ 2DX (indefinite integral of square a plus square x under the root sign)

This problem uses triangle to replace yuan
It uses an integral
∫ secu du = ln|secu+tanu| +C
This integral is also a common integral
This result should be remembered
You can use it directly when doing questions
Satisfied please praise o (∩∩) o

Find 1 + x ^ 2DX under the X radical of indefinite integral

Answer:
∫ x/√(1+x^2) dx
=(1/2) ∫ [1/√(1+x^2)] d(x^2)
=(1/2) ∫ (1+x^2)^(-1/2) d(x^2+1)
=√(1+x^2)+C

Solving indefinite integral ∫ 1 / (2x-3) ^ 2DX by substitution method