finding the antiderivative

[carolyn222]

I have two problems I'm having a hard time with I'm just so lost... the first is (5x+2) / (x^(1/3)) , y(1) = 0 .. as of right now I have gotton to 1/3(15x+6)/x^(1/3) but I think I'm doing this completely wrong. I know it should be easy but I feel so stupid right now.. the other is [(8x^2)*(e^(2x))-4x] / x^2 , y(1)=1 which I have noooo idea how to solve at all.. I had a hard time with derivatives and obviously since this is like the inverse it doesn't help.

 

[carolyn222]

okay I think the first answer should come out to -15x^(5/3)+(3/2)x^(4/3) + 33/2 .... as far as the second one can I cancel out the x^2 in the first part of the top function and leave it with 8e^(2x) - 4ln absolute value of x ???

 

[carolyn222]

my calculus teacher has that as part of a homework to solve I don't understand why she would do that if we couldn't solve it .... are you sure??? (5x+2) / the cubed rt of x

 

[mmm4444bot]

My integration skills faded away some years ago, but I can offer the following antiderivatives calculated by machine, so that you have something to verify.

3x^(5/3) + 3x^(2/3) + C

4 e^(2x) - 4 ln(2x) + C

 

[carolyn222]

okay I think i should be able to figure out the steps in between then from that first break down. thank you!

 

[mmm4444bot]

… as far as the second one can I cancel out the x^2 in the first part of the top function and leave it with 8e^(2x) - 4ln absolute value of x ???

Yes, since each term in the numerator contains some factor of x, you can cancel these with the denominator.

However, I'm not sure how you got 4 ln(|x|) for the second term (after dividing 4x by x^2). That term should be 4/x.

In other words, after cancelling, the integrand becomes:

8 e^(2x) - 4x^(-1)

 

[mmm4444bot]

Hmmm. Something is not right, and I'm not even sure if it matters.

I used the machine to confirm that the antiderivatives are the same, for the following integrands.

[8 x^2 e^(2x) - 4x]/x^2

8 e^(2x) - 4x^(-1)

Respectively, the machine gives the following.

4 e^(2x) - 4 ln(2x) + C

4 e^(2x) - 4 ln(x) + C

I do not understand why there is a discrepancy; however, differentiating both of these antiderivatives yields the same result. In other words, their derivatives are both the same function given in your original exercise.

(I really have no business joining this discussion because I've forgotten too much. I should have kept my mouth shut.)

 

[galactus]

If I may put in my 2 cents.

\(\displaystyle \int\frac{5x+2}{\sqrt[3]{x}}dx\)

\(\displaystyle =5\int x^{\frac{2}{3}}dx+2\int x^{\frac{-1}{3}}dx\)

\(\displaystyle =3x^{\frac{5}{3}}+3x^{\frac{2}{3}}\)

\(\displaystyle =\boxed{3x^{\frac{2}{3}}(x+1)+C}\)

Now, use your initial conditions to find C.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

\(\displaystyle \int\frac{8x^{2}e^{2x}-4x}{x^{2}}dx\)

\(\displaystyle =8\int e^{2x}dx-4\int\frac{1}{x}dx\)

For the first one, let \(\displaystyle u=2x, \;\ \frac{du}{2}=dx\)

\(\displaystyle 4\int e^{u}du=4e^{u}\Rightarrow 4e^{2x}\)

The right one is just 4ln(x)

So, we have: \(\displaystyle 4e^{2x}+4ln(x)+C\)

\(\displaystyle \boxed{4(e^{2x}+ln(x))+C}\)

Now, use your initial conditions to find C.

 

[Deleted member 4993]

If I may put in my 2 cents.

\(\displaystyle \int\frac{8x^{2}e^{2x}-4x}{x^{2}}dx\)

\(\displaystyle =8\int e^{2x}dx-4\int\frac{1}{x}dx\)

For the first one, let \(\displaystyle u=2x, \;\ \frac{du}{2}=dx\)

\(\displaystyle 4\int e^{u}du=4e^{u}\Rightarrow 4e^{2x}\)

The right one is just 4ln(x)

So, we have: \(\displaystyle 4e^{2x}+4ln(x)+C\)

\(\displaystyle \boxed{4(e^{2x}+ln(x))+C}\)

That +C is a hanger on I mostly do not worry about. Unless your teacher is anal, then throw it in there.

But in this case "C" is important - because the original problem gave a condition y(1) = 1

then

y (1)= 4(e^2 + ln1) + C = 1 ? C = 1 - 4e^2

\(\displaystyle \boxed {y(x) \, = \, 4[e^{2x} \, - e^2 \, + \, ln(x)] \, + \, 1}\)

 

[galactus]

Yes, I noticed those afterwards. That is why I edited my post.

 

[carolyn222]

yay I almost got the second one right by myself today if that answer is right