# Calculus Integral Problems

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• Mar 15th 2009, 03:06 PM
defjammer91
Calculus Integral Problems
I need help with two calculus problems. I know that I need to use the second fundamental theorem of calculus for the 1st one, but I have no idea how to do the second one.

1) d/dx integral of ln(t) dt bounded from lower bound (2) and upper bound (x)

2) In decomposing (3(x^2)-2x+19)/(x-2)((x^2)+5) by the method of partial fractions, one of the fractions obtained is:

A) 2/x-2 B) 3/x-2 C) 4/x-2 D) 9/x-2 E) 27/x-2
• Mar 15th 2009, 03:24 PM
Soroban
Hello, defjammer91!

Quote:

I have no idea how to do the second one.
You've never ever had a lesson on Partial Fractions?
Then who assigned this problem to you?

2) In decomposing $\displaystyle \frac{3x^2-2x+19}{(x-2)(x^2+5)}$ by the method of partial fractions,

one of the fractions obtained is:

. . $\displaystyle (A)\;\frac{2}{x-2} \qquad(B)\;\frac{3}{x-2}\qquad(C)\;\frac{4}{x-2}\qquad(D)\;\frac{9}{x-2}\qquad(E)\;\frac{27}{x-2}$

We have: .$\displaystyle \frac{3x^2-2x+19}{(x-2)(x^2+5)} \;=\;\frac{A}{x-2} + \frac{Bx}{x^2+5} + \frac{C}{x^2+5}$

Then: .$\displaystyle 3x^2-2x+19 \;=\;A(x^2+5) + Bx(x-2) + C(x-2)$

Let $\displaystyle x = 2\!:\;\;12-4+19 \:=\:A(9) + B(2)(0) + C(0) \quad\Rightarrow\quad 9A \:=\:27\quad\Rightarrow\quad A = 3$

Hence, the first fraction is: .$\displaystyle \frac{3}{x-2}$ . . . . answer (B)

• Mar 15th 2009, 03:39 PM
rocsfinest762
My math teacher assigned that same problem as extra credit, but we never learned how to do that.
• Mar 15th 2009, 03:43 PM
defjammer91
I wasn't there when the teacher taught that subject.