Question

1)ʃ(x/2-1)^5dx ;2)ʃ1/(2x-1)^3*dx ;3)ʃ1/(3x+1)^4*dx.​

Answer 1

Ответ:

f(x) = ∫ ( 1/3 x² -2x + 5) dx

f(x)= 1/3. 1/3. x³ - 2. 1/2. x² + 5x + c

f(x)= 1/9 x³ - x² + 5x + c

f(0) = 5

1/9(0)³- (0)² + 5(0) + c = 5

c= 5

f(x) = 1/9 x³ - x² + 5x + 5

Answer 2

I see that you're looking to integrate three different expressions with respect to their respective variables. Let's solve each of these integrals step by step:

1. ∫(x/2 - 1)^5 dx: Let's start by expanding the expression inside the integral:

(x/2 - 1)^5 = x^5/32 - 5x^4/32 + 10x^3/32 - 10x^2/32 + 5x/32 - 1

Now, integrate each term separately:

∫x^5/32 dx - ∫5x^4/32 dx + ∫10x^3/32 dx - ∫10x^2/32 dx + ∫5x/32 dx - ∫1 dx

(1/32) * ∫x^5 dx - (5/32) * ∫x^4 dx + (10/32) * ∫x^3 dx - (10/32) * ∫x^2 dx + (5/32) * ∫x dx - ∫dx

Now integrate each term using the power rule:

(1/32) * (x^6/6) - (5/32) * (x^5/5) + (10/32) * (x^4/4) - (10/32) * (x^3/3) + (5/32) * (x^2/2) - x + C

Simplify and combine the terms:

x^6/192 - x^5/32 + 5x^4/128 - 5x^3/96 + 5x^2/64 - x + C

2. ∫1/(2x - 1)^3 dx: Start by making a substitution: let u = 2x - 1, then du = 2 dx.

The integral becomes:

(1/2) ∫du/u^3

Now integrate with respect to u:

(1/2) * (-1/2u^2) + C

Simplify and substitute back u = 2x - 1:

-1/(4(2x - 1)^2) + C

3. ∫1/(3x + 1)^4 dx: Start by making a substitution: let u = 3x + 1, then du = 3 dx.

The integral becomes:

(1/3) ∫du/u^4

Now integrate with respect to u:

(1/3) * (-1/3u^3) + C

Simplify and substitute back u = 3x + 1:

-1/(9(3x + 1)^3) + C

So, the integrated expressions are:

1. x^6/192 - x^5/32 + 5x^4/128 - 5x^3/96 + 5x^2/64 - x + C1
2. -1/(4(2x - 1)^2) + C2
3. -1/(9(3x + 1)^3) + C3

Where C1, C2, and C3 are integration constants.