Function of Function Rule? (1 Viewer)

[Avenger6]

I've been struggling with this one for a while now, any help would be greatly appreciated .

Find the indefinite integral of:

F 1/[2(4x-5)^3] dx

 

[YannY]

I've been struggling with this one for a while now, any help would be greatly appreciated .

Find the indefinite integral of:

F 1/[2(4x-5)^3] dx

Function of a function rule yes.

But in integration lets just call it subsitution method.

Let u=4x-5 dx=du/4

now subsitute these two values in the integral: (integrand i.e f sign)
1/(2u^3).du/4
This can be simplified to du/(8u^3)

Now integrate this normal to get -1/(16u^2) + C

Now subsitute back the u=4x-5

to get -1/(16(4x-5)^2) + C

Lets differentiate to make sure

d/dx [-1/(16(4x-5)^2)]
=d/dx[(-1/16) (4x-5)^-2]
=1/2 (4x-5)^-3
=1/[2(4x-5)^3]

So we're correct! =]

 

[Avenger6]

Thankyou for your response. Im following up until the point where you integrate the normal to get -1/(16u^2) + C. I thought you added 1 to n when integrating, so shouldn't it become du/8u^4/2 which comes to 1/2u^4??? Help is once more, greatly appreciated.

 

[Mark576]

du/(8u³) = (1/8) . u^-3 . du = (1/8).(-1/2) . u^-2 + c = -1/(16u²) + c

 

[ssglain]

Thankyou for your response. Im following up until the point where you integrate the normal to get -1/(16u^2) + C. I thought you added 1 to n when integrating, so shouldn't it become du/8u^4/2 which comes to 1/2u^4??? Help is once more, greatly appreciated.

Yes, you add 1 to n when integrating - but notice that the integral is 1 OVER 8u³? In order to integrate, it must be written in the standard xⁿ form. If we do that, it becomes (1/8)u^-3. Now if you proceed to carry out integration by adding 1 to the index and dividing by the new index, you will get (-1/16)u^-2, which is -1/(16u²).

 

[Trebla]

A general rule that might help:
∫ [f(x)]ⁿ.f'(x) dx = [f(x)]^{n + 1} / (n + 1) + c
This is easily verified by differentiating the primitive.

 

[Avenger6]

Thanks for all the help.