# Math Help - Antiderivative problem

1. ## Antiderivative problem

The marginal cost of producing the Xth role of film is given by 5+1/(X+1)^2. The total cost to produce one roll is $1000. find the total cost function. to take the antiderivative of 5+1/(X+1)^2 I replaced X+1 with R and wrote (5+1)(R^-2). I took the antiderivative of this : (5+1)(R^-1)/-1 I replaced R with X+1 : (5+1)/(X+1) What am I doing wrong? 2. Originally Posted by Kenneth The marginal cost of producing the Xth role of film is given by 5+1/(X+1)^2. The total cost to produce one roll is$1000. find the total cost function.

to take the antiderivative of 5+1/(X+1)^2 I replaced X+1 with R and wrote

(5+1)(R^-2). I took the antiderivative of this :

(5+1)(R^-1)/-1

I replaced R with X+1 :

(5+1)/(X+1)

What am I doing wrong?
5 + 1/(X+1)^2 = 5 + (x + 1)^-2

int{5 + (x + 1)^-2}dx = 5x - (x + 1)^-1 + C

the last step i could do only because the power of x is 1. if we had x^2 + 1 we couldn't do that

3. Thanks. Could you help me with a similiar problem.

int{t(t^2+1)^4 + t} dx

I substitued u for (t^2+1) :

int{t(u)^4 + t/2t} du
= int{(u)^4 + t/2} du

Then I replaced t with (u-1)^1/2 :

int{(u)^4 + (u-1)^1/2/2} du

(u^5)/10 + (u^3/2)/3
= (t^2+1)^5/10 + t^3/3

is this correct?

4. Originally Posted by Kenneth
Thanks. Could you help me with a similiar problem.

int{t(t^2+1)^4 + t} dx

I substitued u for (t^2+1) :

int{t(u)^4 + t/2t} du
= int{(u)^4 + t/2} du

Then I replaced t with (u-1)^1/2 :

int{(u)^4 + (u-1)^1/2/2} du

(u^5)/10 + (u^3/2)/3
= (t^2+1)^5/10 + t^3/3

is this correct?
half of your answer is right, not sure what happened with the other half. i see you tried to use the substitution on the lone t (which is where you made a mistake), it can work, but it makes life complicated. here's what you do: split the integral into two, and work on each piece separately. (by the way, it should be dt not dx)

int{t(t^2+1)^4 + t} dt
= int{t(t^2+1)^4} dt + int{t}dt
for int{t(t^2+1)^4} dt
let u = t^2 + 1
=> du = 2t dt
=> (1/2)du = t dt
so our integral becomes:
(1/2)int{u^4}du
= (1/2)(1/5)u^5 + C
= (1/10)(t^2 + 1)^5 + C

so int{t(t^2+1)^4} dt + int{t}dt = (1/10)(t^2 + 1)^5 + (1/2)t^2

for the last term, you had t^3/3, should be (t^2)/2