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# cal 2 integral question?

Please help! I got the answer ln(5)- ln(4)+3ln(2).

I plugged in the question into an integral website and it got ln(5)-ln(4)-2ln(2). Don't know where I went wrong!

### 5 Answers

- SlowfingerLv 65 months agoFavorite Answer
husoski got that right.

I am going to check that, just as a mental gymnastics, in different order of operations.

₀∫¹(x-8)/(x²-7x+10) dx =

substitute immediately

u=2x-7 (yes, that's derivative of the denominator), => x=(u+7)/2, du=2dx

new lower bound u=2*0-7=-7 upper bound u=2*1-7=-5

= (1/2) ₋₇∫⁻⁵ ((u+7)/2-8)/((u+7)²/4-7(u+7)/2+10) du

= (1/4) ₋₇∫⁻⁵ ((u+7)-16)/((u+7)²/4-7(u+7)/2+10) du

= ₋₇∫⁻⁵((u+7)-16)/((u+7)²-14(u+7)+40) du =

= ₋₇∫⁻⁵(u-9)/(u²+14u+49-14u-14*7+40) du =

= ₋₇∫⁻⁵(u-9)/(u²-9) du =

partial fractions

= ₋₇∫⁻⁵(2/(u+3) - 1/(u-3)) du =

integrate term by term

= 2₋₇∫⁻⁵1/(u+3) du - ₋₇∫⁻⁵1/(u-3) du

= 2ln|u+3|₋₇⁻⁵ - ln|u-3|₋₇⁻⁵

= 2 [ln|-5+3| - ln|-7+3|] - [ln|-5-3| - ln|-7-3| ]

= 2 [ln|-2| - ln|-4|] - [ln|-8| - ln|-10| ]

= 2 [ln(2)-ln(4)] - [ln(8)-ln(10) ]

= -1.16315...

Edit:

The result can be shortened using basic operations with logarythms

2(ln(2)-ln(4)) - (ln(8)-ln(10)) =

= 2 ln(2/4) - ln(8/10)

= 2 ln(1/2) - ln(4/5)

= ln((1/2)²) - ln(4/5)

= ln((1/4)/(4/5))

= ln(5/16)

- husoskiLv 75 months ago
You used partial fractions, right? The denominator factors as (x - 2)(x - 5), so you want to find constants A, B so that:

(x - 8) / (x² - 7x + 10) = A / (x - 2) + B / (x - 5)

Multiply by (x² - 7x + 10) = (x - 2)(x - 5):

x - 8 = (x - 5)A + (x - 2)B = (A + B)x - 5A - 2B

The coefficients on x and the constant terms must be independently equal:

1 = A + B

-8 = -5A - 2B

Negate the 2nd eq. and substitute B = 1 - A (from the 1st eq.):

8 = 5A + 2B = 5A + 2(1 - A)

6 = 5A - 2A = 3A

A = 6/3 = 2

B = 1 - A = -1

(x - 8) / (x² - 7x + 10) = 2/(x - 2) - 1/(x - 5)

∫ (x - 8) / (x² - 7x + 10) dx = 2 ln |x - 2| - ln |x - 5| + C

The definite integral follows from that.

I'd simplify the result, though:

ln 5 - ln 4 - 2 ln 2 = ln [ 5 / (4 * 2²) ] = ln (5/16)

...or ln 5 - 4 ln 2 if you only want logs of small integers.

- rotchmLv 75 months ago
Apply what I already told you in:

https://ca.answers.yahoo.com/question/index?qid=20...

Anything in there you still need help with before we tackle the current question?

This is how you will *learn*!

- AshLv 75 months ago
₀∫¹ (x-8)/(x²-7x+10) dx

= ₀∫¹ (x-8)/(x-5)(x-2) dx

(x-8)/(x-5)(x-2) = A/(x-5) + B/(x-2)

(x-8)/(x-5)(x-2) = [A(x-2) + B(x-5)]/(x-5)(x-2)

(x-8) = (Ax - 2A + Bx - 5B)

(x-8) = (A+B)x - (2A + 5B)

then we get

A+B = 1 .....(1) and

2A+5B = 8 ...(2)

Multiple (1) by 2 and subtract from (2)

2A+5B - 2(A+B) = 8 - 2

2A+5B - 2A - 2B = 6

3B = 6

B = 2

plug in (1)

A + 2 = 1

A = -1

Then the integration becomes

= ₀∫¹ -1/(x-5) + 2/(x-2) dx

= -₀∫¹1/(x-5) dx + 2₀∫¹1/(x-2) dx

Now we know ∫1/(x+a) = 1/a ln|ax+b| + C , then we get

= [-ln|x-5| + 2ln|x-2|] ₀|¹

= (-ln|1-5| + 2ln|1-2|) - (-ln|0-5| + 2ln|0-2|)

= -ln4 + 2ln1 + ln5 - 2ln2

= ln5 - ln4 - 2 ln2

we can further simplify as

= ln5 - ln2² - 2 ln2

= ln5 - 2ln2 - 2ln2

= ln5 - 4ln2

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