1. Integrals using U-Substitution

Can anyone check if I'm doing these correct

1. $\displaystyle \int{\frac{(lnx)^2}{x}dx}$

$\displaystyle U = lnx , DU = \frac{dx}{x}$

$\displaystyle \int{u}{du}$

$\displaystyle = \frac{1}{2}(u^2)$

$\displaystyle = \frac {1}{2}(lnx)^2$

2. $\displaystyle \int\frac{x}{(x^2+2)^2}dx$

$\displaystyle U = x^2 + 1 , \frac{1}{2}DU = xdx$

$\displaystyle = \frac{1}{1}\int\frac{1}{u^2}du$

$\displaystyle = \frac{1}{2}ln(u^2)$

$\displaystyle = \frac{1}{2}ln|(x^2)+1|$

I know I made mistakes I just don't know where.

2. You made a mistake in #1:

after you substitute u = ln(x) you should get:

$\displaystyle \int u^2 du$

And in #2 you should make $\displaystyle u = x^2 + 2$.

3. Originally Posted by TrevorP
You made a mistake in #1:

after you substitute u = ln(x) you should get:

$\displaystyle \int u^2 du$

And in #2 you should make $\displaystyle u = x^2 + 2$.
Yes, and in the second one, you would then be integrating $\displaystyle u^{-2}$ which is $\displaystyle -u^{-1}$

4. Originally Posted by FalconPUNCH!
$\displaystyle = \frac{1}{1}\int\frac{1}{u^2}du$

$\displaystyle = \frac{1}{2}ln(u^2)$
This is a common mistake.

It is $\displaystyle \int {\frac{1} {u}\,du} = \ln \left| u \right| + k.$ For $\displaystyle \int {\frac{1} {{u^2 }}\,du}$ write the integrand as $\displaystyle u^{-2}$ and apply the power rule.

5. Originally Posted by FalconPUNCH!
Can anyone check if I'm doing these correct

1. $\displaystyle \int{\frac{(lnx)^2}{x}dx}$

$\displaystyle U = lnx , DU = \frac{dx}{x}$

$\displaystyle \int{u}{du}$

$\displaystyle = \frac{1}{2}(u^2)$

$\displaystyle = \frac {1}{2}(lnx)^2$

2. $\displaystyle \int\frac{x}{(x^2+2)^2}dx$

$\displaystyle U = x^2 + 1 , \frac{1}{2}DU = xdx$

$\displaystyle = \frac{1}{1}\int\frac{1}{u^2}du$

$\displaystyle = \frac{1}{2}ln(u^2)$

$\displaystyle = \frac{1}{2}ln|(x^2)+1|$

I know I made mistakes I just don't know where.
Most of what needs to be said has been said and said well.

My 2 cents is that if the original questions asked for the integral, the arbitrary constant of integration must be included in the answer.

6. Thanks guys. Sorry I didn't respond back until today. Well I have more problems which are harder and I don't really know where to start with some of them.

1. $\displaystyle \int\frac{x^2}{\sqrt{1-x}}dx$

For this one I don't know here to begina

2. $\displaystyle \frac{sinx}{1+cos^{2} x} dx$

3. $\displaystyle \frac{sin2x}{1+cos^{2}x}dx$

For these two I was never that good in trig. so it's hard for me to understand where to start.

4.$\displaystyle \int_{0}^{1} (x)\sqrt{x-1}dx$

I'm not really looking for answers but I am looking for how to start these problems. Thank you guys

7. Use u-substitution for the first one:

$\displaystyle u=1-x$
$\displaystyle du=-dx$

$\displaystyle x=1-u$
$\displaystyle x^2=(1-u)^2 \Rightarrow u^2 - 2u + 1$

Your integrand then becomes: $\displaystyle \frac{-u^2+2u-1}{\sqrt{u}}$

Separate each term into its own fraction and apply the power rules for integration to each.

8. Originally Posted by FalconPUNCH!

1. $\displaystyle \int\frac{x^2}{\sqrt{1-x}}dx$
Substitute $\displaystyle u^2=1-x.$

Originally Posted by FalconPUNCH!
2. $\displaystyle \frac{sinx}{1+cos^{2} x} dx$
This is an arctangent. First substitute $\displaystyle u=\cos x.$

Originally Posted by FalconPUNCH!
3. $\displaystyle \frac{sin2x}{1+cos^{2}x}dx$
This was posted before, you just need to note that

$\displaystyle \left(1+ {\cos ^2 x} \right)' = - 2\sin x\cos x = - \sin 2x.$

Originally Posted by FalconPUNCH!
4.$\displaystyle \int_{0}^{1} (x)\sqrt{x-1}dx$
Substitute $\displaystyle u^2=x-1.$ (Do you know Barrow's Rule? When makin' a change of variables you need to get the new integration limits for $\displaystyle u.$)

9. Thank you all for your help. I will try to solve them and come back tomorrow or after tomorrow with my results.

Edit: I do not know Barrow's Rule.

10. Originally Posted by FalconPUNCH!

3. $\displaystyle \frac{sin2x}{1+cos^{2}x}dx$

For this one is u = sin2x?

11. Originally Posted by FalconPUNCH!
For this one is u = sin2x?
No, I would say not. There is a trig identity that says sin2x = 2sinxcosx.

If you fill that in then set u = cos x, you should get somewhere.

12. Ok I sort of understood. I'm not to good with trig but here's what I did

$\displaystyle \int\frac{2sinxcosx}{1+cos^{2}x}$

$\displaystyle u = cos x ; du = -sinxdx$

$\displaystyle -2\int\frac{u}{1+u^2}du$

$\displaystyle -2tan^{-1}(cos^{2}x) + c$

13. Yes the only problem I see with that is that you should have -2Arctan(Arccos(x))

14. Originally Posted by TrevorP
Yes the only problem I see with that is that you should have -2Arctan(Arccos(x))

Sorry, I think that's wrong.

You had $\displaystyle \int \frac{2sinxcosx}{1+cos^2x}dx$

Integrating this gives $\displaystyle -ln(1+cos^2x)+C$

15. Originally Posted by a tutor
Sorry, I think that's wrong.

You had $\displaystyle \int \frac{2sinxcosx}{1+cos^2x}dx$

Integrating this gives $\displaystyle -ln(1+cos^2x)+C$