antidifferentiation help?

Question

I have 2 questions :)
1. Antidifferentiate (X+1)^2 / X
2. Antidifferentiate 2X+1/ X+1

Please show working out, Thanks!

Sorry, I also have another question!
3. Given that dy/dx = 3/ X-2 and y=10 when X=0, Find an expression for y in terms of X.

So I antidifferentiated it an got 3log IX-2I +c, I know I have to sub in y=10 and X=0, but I'm having trouble re-arranging it to get the answer. Help anyone??

Answer 1

1)
dy/dx = (x + 1)²/x

Therefore,
y = ∫(x + 1)²/x dx

y = ∫(x² + 2x + 1)/x dx
y = ∫(x + 2 + 1/x) dx
y = 1/2 x² + 2x + ln x + C

2)
y = ∫(2x + 1)/(x + 1) dx

Let u = x + 1
x = u - 1
dx = du
y = ∫(2(u - 1) + 1)/u du
y = ∫(2u - 2 + 1)/u du
y = ∫(2u - 1)/u du
y = ∫(2 - 1/u) du
y = 2u - ln u + C
y = 2(x + 1) - ln (x + 1) + C

Answer 2

First One ((x+1)^2)/x (I hope thats the one)
= (x^2 + 2x + 1)/x
= x + 2 + 1/x
So Integrating it we will get
(x^2)/2 + 2x + ln|x| + c

Second one = (2x+1)/(x+1) (or is it 2x + (1/x) + 1)
(2x+1)/(x+1) = 2 - (1/(x+1))
So integerating we get
2x - ln|x+1| + c


Third question dy/dx = 3/(x-2)
y' = 3/(x-2)
y = 3ln|x-2| + c
for x=0 y=10
10 = 3ln2 + c c = 10 - 3ln2
y = 3ln(x-2) + 10 - 3ln(2)
y = 10 + 3ln(|2-x|/2))

Answer 3

if u mean integration, this is the answer

(x+1)^2/x = (x^2 + 2x + 1)/x = x + 2 + 1/x

integral (x+1)^2/x = integral [x + 2 + 1/x] = integral x + integarl 2 + integral 1/x <== that's easy to integrate

the second one:

integral 2X+1/ X+1 = integral 2x/(x+1) + integral 1/(x+1)

integral 1/(x+1) <== i hope u know it

integral 2xdx/(x+1) = integarl 2(u-1)du/u = integarl 2 du + integral -2du/u <== easy to solve

THE THIRD:

3ln2 + c =10 ==> c = 10 - 3ln2 , that set
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if u need more help, email me. when i wake up tomorrow, i will answer it.

Answer 4

You'll have to do the first two as integration by parts.
Integral v∙du = uv - integral u∙dv

For the 3'rd one, set y = 3∙log |x-2| +c then put in your x and y values and solve it for c


Doug

Answer 5

1) a. you must expand numerator x^2 + 2x +1
b. you divide by x ... resulting x + 2 + x^(-1)
c. you apply anti-differentiation rules to each part
x^2/2 + 2x + ln(abs(x)) + C

2) Supposing you wrote correct the question, then you must just apply anti-differentiation rules to each part
x^2 + ln(abs(x)) + x + C

2) Supposing you want to say (2x+1)/(x+1)
a. divide numerator by denominator, obtaining 2 - 1/(x+1)
b. apply anti-differentiation rules 2x - ln(abs(x+1)) + C