# More integration:evaluation

• Jun 16th 2008, 08:22 PM
deragon999
More integration:evaluation
Just put a yes or no as to my answer =P, c is a constant, sorry about format and squareroots

1) integrate: (x^4-2x^3+5x+1)/(x^4) dx

I got: x -5/(2x^3) -3/(x^3) -2 +c

2) integrate: sinx/(squareroot of cosx) dx

I got: -0.5(squareroot of cosx) +c

3) integrate (e^x)squareroot(e^x+4) dx
I got: 0.5(squareroot e^x+4) +c

4) integrate xe^(x^2+1) dx

I got: 0.5e^(x^2+1) +c

5) integrate 1/((squareroot x)((squareroot x) +1)^2) dx
It is 1 divided by (rootx)(rootx plus 1)squared

anyways..I got: -2/(squareroot x +1) +c

LAST ONE =P

6) integrate (x^2)/(1+x^6) done through two substitutions

let u=x^3 , then let u=root3 tan theta

In the end i got: (squareroot 3)*arctan((x^3)/(squareroot 3)) +c

arctan is tan^-1 for anyone maths illiterate reading this.

• Jun 16th 2008, 08:43 PM
Jhevon
Quote:

Originally Posted by deragon999
Just put a yes or no as to my answer =P, c is a constant, sorry about format and squareroots

1) integrate: (x^4-2x^3+5x+1)/(x^4) dx

I got: x -5/(2x^3) -3/(x^3) -2 +c

Incorrect:

note that $\displaystyle \frac {x^4 - 2x^3 + 5x + 1}{x^4} = \frac {x^4}{x^4} - \frac {2x^3}{x^4} + \frac {5x}{x^4} + \frac 1{x^4} = 1 - \frac 2x + 5x^{-3} + x^{-4}$

Quote:

2) integrate: sinx/(squareroot of cosx) dx

I got: -0.5(squareroot of cosx) +c
Incorrect: your constant is off. check again (you can differentiate the answer to see that you are wrong)

Quote:

3) integrate (e^x)squareroot(e^x+4) dx
I got: 0.5(squareroot e^x+4) +c
Incorrect again: are you sure you didn't mean $\displaystyle \int \frac {e^x}{\sqrt{e^x + 4}}~dx$? your answer would still be wrong, but you would have been closer

Quote:

4) integrate xe^(x^2+1) dx

I got: 0.5e^(x^2+1) +c
correct (Clapping)

Quote:

5) integrate 1/((squareroot x)((squareroot x) +1)^2) dx
It is 1 divided by (rootx)(rootx plus 1)squared

anyways..I got: -2/(squareroot x +1) +c
Correct (Clapping)

Quote:

LAST ONE =P
Okie dokie :p

Quote:

6) integrate (x^2)/(1+x^6) done through two substitutions

let u=x^3 , then let u=root3 tan theta

In the end i got: (squareroot 3)*arctan((x^3)/(squareroot 3)) +c

arctan is tan^-1 for anyone maths illiterate reading this.

incorrect:

just a substitution of u = x^3 will do. no second substitution is needed, and trig sub is definitely not needed
• Jun 16th 2008, 08:57 PM
Mathstud28
Quote:

Originally Posted by deragon999
Just put a yes or no as to my answer =P, c is a constant, sorry about format and squareroots

1) integrate: (x^4-2x^3+5x+1)/(x^4) dx

I got: x -5/(2x^3) -3/(x^3) -2 +c

2) integrate: sinx/(squareroot of cosx) dx

I got: -0.5(squareroot of cosx) +c

3) integrate (e^x)squareroot(e^x+4) dx
I got: 0.5(squareroot e^x+4) +c

4) integrate xe^(x^2+1) dx

I got: 0.5e^(x^2+1) +c

5) integrate 1/((squareroot x)((squareroot x) +1)^2) dx
It is 1 divided by (rootx)(rootx plus 1)squared

anyways..I got: -2/(squareroot x +1) +c

LAST ONE =P

6) integrate (x^2)/(1+x^6) done through two substitutions

let u=x^3 , then let u=root3 tan theta

In the end i got: (squareroot 3)*arctan((x^3)/(squareroot 3)) +c

arctan is tan^-1 for anyone maths illiterate reading this.

I am feeling good tonight, I will do all of these for didactic purposes.

1)$\displaystyle \int\frac{x^4-2x^3+5x+1}{x^4}dx=\int\bigg[x-\frac{2}{x}+\frac{5}{x^3}+\frac{1}{x^4}\bigg]dx=x-2\ln(x)-\frac{5}{x^2}-\frac{1}{3x^3}+C$

2)$\displaystyle \int\frac{\sin(x)}{\sqrt{\cos(x)}}dx=-\int\frac{-\sin(x)}{\sqrt{\cos(x)}}dx$

Now let $\displaystyle \vartheta=\cos(x)\Rightarrow{d\vartheta=-\sin(x)}$

So we have

$\displaystyle -\int\frac{d\vartheta}{\sqrt{\vartheta}}=-2\sqrt{\vartheta}+C$

Backsubbing

$\displaystyle -2\sqrt{\cos(x)}+C$

3)$\displaystyle \int{e^x\sqrt{e^x+4}dx}$

Let $\displaystyle \varphi=e^x\Rightarrow{d\varphi=e^x}$

Giving us

$\displaystyle \int\sqrt{\varphi+4}d\varphi=\frac{2}{3}(\varphi+4 )^{\frac{3}{2}}+C$

backsubbing we get

$\displaystyle \frac{2}{3}\left(e^x+4\right)^{\frac{3}{2}}+C$

4) $\displaystyle \int{xe^{x^2+1}dx}$

Let $\displaystyle \psi=x^2+1\Rightarrow{d\psi=2x}$
So we have

$\displaystyle \frac{1}{2}\int{e^{\psi}d\psi}=\frac{1}{2}e^{\psi} +C$

Backsubbing we get

$\displaystyle \frac{1}{2}e^{x^2+1}+C$

5) $\displaystyle \int\frac{dx}{\sqrt{x}(\sqrt{x}+1)^2}$

Let $\displaystyle \xi=\sqrt{x}\Rightarrow\xi^2=x\Rightarrow{2\xi=dx}$

So we have

$\displaystyle 2\int\frac{\xi}{\xi(1+\xi)^2}=2\int\frac{d\xi}{(1+ \xi)^2}=\frac{-2}{1+\xi}+C$

Backsubbing we get

$\displaystyle \frac{-2}{1+\sqrt{x}}+C$

6) $\displaystyle \int\frac{x^2}{1+x^6}dx$

Let $\displaystyle \zeta=x^3\Rightarrow{d\zeta=3x^2}$
So now we have

$\displaystyle \frac{1}{3}\int\frac{d\zeta}{1+\zeta^2}=\frac{1}{3 }\arctan(\zeta)+C$

Now backsubbing we get

$\displaystyle \frac{1}{3}\arctan(x^3)+C$
• Jun 16th 2008, 09:41 PM
deragon999
Ok thanks i think i get all that, funny symbols and all =P, i just differentiated instead of integrating at some points..doh..and made a few minor errors that totally screwed me over...hmmm
• Jun 16th 2008, 09:45 PM
Mathstud28
Quote:

Originally Posted by deragon999
.. funny symbols and all =P..

You say abc I say $\displaystyle \alpha\beta\gamma$
• Jun 16th 2008, 09:54 PM
deragon999
Well actually when the mood takes me i say a i u e o ka ki ku ke ko...

Btw could you please check out my post just before this one..more integration gone wrong i guess...
• Jun 16th 2008, 09:56 PM
Mathstud28
Quote:

Originally Posted by deragon999
Well actually when the mood takes me i say a i u e o ka ki ku ke ko...

Btw could you please check out my post just before this one..more integration gone wrong i guess...

What language is that?
• Jun 16th 2008, 09:57 PM
deragon999
Jap..tho i hardly speak it since i stopped learning 2 years ago

nihongo o wakrimasu ka
• Jun 16th 2008, 09:58 PM
Mathstud28
Quote:

Originally Posted by deragon999
Jap..tho i hardly speak it since i stopped learning 2 years ago

Oh...ok. Dont feel bad, I have not spoken any French this summer
• Jun 16th 2008, 10:00 PM
deragon999
so thats a big yes to you cheking out my other post =P...all simple stuff i hope