Intergral of (X)/(X+10) dx. From 0 to 5?

Question

We are learning U-Sub, Integration by Parts, and Partial Fractions. So using any of the above to answer the question.
Thank you.

Answer 1

This is a partial fractions type question.
Rule of thumb: Whenever the degree of the numerator is greater than or equal to the degree of the denominator, you want to use long division to change the question. In the above case, you'll want to divide X + 10 into X. Due to the limitations of text you'll have to bear with me here:
1
--------------------
X+10 | X+0
x+10
-------
-10
So X/(X+10) = -10/(x+10) + 1

Now, you integrate that.

Integral (0,5) [X/(X+10)]dx =
Integral (0,5) [-10/(x+10) + 1] dx

Split it up into two integrals, to get

Integral (0,5) [-10/(x+10)] dx - Integral (0,5) [1]dx

Pull out any constants. The second integral is trivial to solve

-10 * Integral (0,5) [1/(x+10)]dx - Integral (0,5) [1] dx

Through simple substitution, you'll learn that the integral of 1/(x+10) is equal to ln|x+10|. The second integral is trivial. So we get
5 5
-10 * (ln|x+10)] | - x |
0 0
=

-10 * (ln|5+10| - ln|0+10|) - (5 - 0)
-10 * (ln|15| - ln|10|) - 5

Remember the log rule when dealing with subtraction.

-10 * ln(15/10) - 5
-10 * ln (3/2) - 5

Answer 2

Express the integrand, x/(x + 10)

(presumably x here; not X, which would be some
other variable) as:

(x + 10 - 10)/(x +10)

(The idea is to get rid of the awkward x on top by
combining it with something that can make it
easily divide out; then correct for the change that
you made to bring the top back to what you
started with. Use similar ideas in future.)

So, doing the dividing, we now have: 1 - 10/(x +10).

You should now recognize 1/(x + 10) as an easily
integrable function. Have at it!

Live long and prosper.

Answer 3

there is a trick in this integral :
for the numerator, just do X = X+10 -10; then separate.
From 0 to 5 :
=>Integral of ( (X+10)/(X+10)dx - Integral of(10 /(X +10)) dX
<=> Integral of dX - 10*Integral (1/(X+10)) dX

note that as X+10> 0 on [0;5] , so you can say that:
Integral[0;5] (1/(X+10)) dX = {ln (X+10)} on [0;5]
(review Integrations about your log functions )
so, you'll get :
(5-0) - 10* {ln (5+10) -ln10} = 5 -10* {ln 15 - ln 10}
= 5 - 10*ln (15/10)
= 5 -10*ln(3/2)

I hope it would help for your future integrals!!

Answer 4

Integral is X - 10ln(X+10)

Which gives 5 - 10ln3

Answer 5

Partial fractions:

x/(x+10) = (x+10-10)/(x+10) = 1 - 10/(x+10)

Antiderivative: x - 10ln(x+10)

Definite int.: 5 - 10ln(1.5)

Answer 6

let u=x+10; then x=u-10 and dx=du

so int_0_5 (x/x+10) dx = int_10_15 (u-10/u)du
=int_10_15 (1-10/u) du = (u-10logu)_10_15
=15-10log15-10+10log10 = 5-10log1.5