I understand that the integral of 1/x is ln x + c where x > 0

but what happens if you integrate 4/x?

Is it 4.ln x + c?

I'm not sure where the 4 fits in to the answer.

but what happens if you integrate 4/x?

Is it 4.ln x + c?

I'm not sure where the 4 fits in to the answer.

by substituting

u = 2x

du = 2dx

dx = du/2

you can get

(1/2) ln(2x) + c

if you differentiate that, you get

(1/2) 2 / 2x = 1/(2x)

however, note that

(1/2) ln(2x) + c = (1/2) (ln(x) + ln(1/2)) + c = (1/2) ln(x) + (1/2) ln(1/2) + c

= (1/2) ln(x) + k

absorbing the term into another arbitrary constant in that case.

§ b f(x) dx = b § f(x) dx

i.e., you can bring the constant factor outside of the integral.

You may have already seen that the integral can be used to calculate the area under a curve. So, if you calculated the area under the function f(x) in some interval, then if you consider another function g(x) = 2 f(x), then g(x) is twice the value of f(x) for any particular x. Then you would expect that the area under the curve g(x) to be twice as much as the area under the curve f(x).

For your indefinite integra, if

§ f(x) dx = h(x) +c

then from above, you would say

§ b f(x) dx = b § f(x) dx = b( h(x) +c ) = b h(x) + b c

where h(x) is just the integral of f(x)

But since b and c are both constants, you can just treat them as a single constant, call it C.

§ b f(x) dx = b h(x) + C

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4 * 1/x

Integrate and you get

4 ln x + k

Here my constant K is arbirtrary and is equal to 4*c.

4 times a constant is just another constant.

Does that help?