help integrating ∫(2x^2 - 5x -2)/x dx in respect to x?

Question

i need to integret these in respect to x can you describe in detail as i dont understand it very well.
1)∫(2x^2 - 5x -2)/x dx
2)∫upper limit 5 lower limit 2 4x^5 - 6x dx
3)∫upper limit 4 lower limit 1 3/x^4 dx

Answer 1

1) rewrite.

(2x^2 - 5x - 2)/x = 2x - 5 + (2/x)
int(2x - 5 + 2/x) = 2(x^2/2) - 5x + 2 * ln |x|

Here you use the power rule, int(x) = x^2/2 and int(1) = x. And you should remember that int(1/x) = ln |x|.

2) Use the power rule.

int(4x^5 - 6x) = 4(x^6/6) - 6(x^2/2) = (2/3)x^6 - 3x^2
Now plug in your bounds.
(2/3)(5)^6 - 3(5)^2 - (2/3)(2)^6 + 3(2)^2
=10311 (I'd use a calculator here, this is messy)

3) Rewrite

3/(x^4) = 3x^(-4)
Now use the power rule.

int(3x^(-4)) = 3(x^(-3)/(-3)) = - x^(-3)
Plug in bounds.
-(4)^(-3) + 1^(-3) = -1/64 + 1 = 63/64

Answer 2

1)
∫(2x^2 - 5x -2)/x dx
= ∫(2x - 5 -2/x) dx
= x^2 - 5x - 2ln|x| + c

2)
∫upper limit 5 lower limit 2 4x^5 - 6x dx
= (4/6)x^6 - 3x^2, upper limit 5 lower limit 2
= 10311

3)
∫upper limit 4 lower limit 1 3/x^4 dx
= -1/x^3, upper limit 4 lower limit 1
= 0.98438

Answer 3

shortcut method; Increase the power and divide everything by new power, however it is a fraction, so flip it and times everything by it. Also have to times by the differential of the brackets. Therefore answer is; (3/2)(5)(5x+2)x^3/2 =15/2(5x+2)x^3/2

Answer 4

Question 1
I = ∫ (2x - 5 - 2/x) dx
I = x² - 5x - 2 log x + C
Question 2
I = ∫ 4x^5 - 6x dx
I = 4 x^6/6 - 3x² between lims 2 to 5
I = [ (4 x 5^6 / 6 ) - 75 ] - [4 x 2^6/6 - 12 ]
I = 10342 - 31
I = 10311 (to nearest whole number)
Question 3
I = ∫3.x^(-4).dx
I = 3.x^(-3) / (-3)
I = - 1 / x³ between limits 1 to 4
I = - [1/64 - 1 ]
I = 63 / 64

Answer 5

seperate the numerator u get.........

=2 xsq /x - 5 x/x - 2/x
Now integrate using standard results as-

integral(1/x)=log x
int(1)=x
int(x)=x^2 /2

so ans is x^2 - 5 x -2log(x)

Answer 6

i cannot teach u integration just in this small space it requires thorogh understanding of concepts and remembering som standard proof results...