4 Integrals (partial fractions and completing the square)

[Stereotypical]

Hello. Please help me with those 4 integrals:

1. \(\displaystyle \int \frac{\mathrm{d}x}{(x^2 + 4x +13)^{\frac{3}{2}}}\)


2. \(\displaystyle \int \frac{x^3\mathrm{d}x}{(x + 1)^2(x-1)}\)

3. \(\displaystyle \int x\sqrt{-8 + 6x - x^2}\mathrm{d}x \)

4. \(\displaystyle \int \frac{3x^2 + 4x + 4 }{x^3 + x}\mathrm{d}x \)

Thanks in advance.

 

[renegade05]

Well what have you tried so far?

Do you know how to complete the square? Do you know the method behind partial fractions?

Show some work so we can see where you are stuck.

 

[galactus]

The first one can indeed be done by completing the square.

\(\displaystyle \int\frac{1}{\sqrt{((x+2)^{2}+9)^{3}}}dx\)

Now, let \(\displaystyle u=x+2, \;\ du=dx\)

Make the subs and you have an integral in terms of u.

Next, make the sub \(\displaystyle u=3tan\theta, \;\ du=3sec^{2}\theta d\theta\)

Making the subs and simplifying causes it to reduce down to a really easy integral.

For #2: partial fractions.

\(\displaystyle \frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{(x+1)^{2}}=x^{3}\)


For #3:

Here is one way to deal with that pesky x on the outside.

\(\displaystyle \int x\sqrt{-x^{2}+6x-8}dx\)

Rewrite it as:

\(\displaystyle \int (x-3)\sqrt{-x^{2}+6x-8}dx+3\int\sqrt{-x^{2}+6x-8}dx\)

\(\displaystyle \int (x-3)\sqrt{1-(x-3)^{2}}dx+3\int\sqrt{1-(x-3)^{2}}dx\)

Now, let \(\displaystyle t=-x^{2}+6x-8, \;\ dt=(-2x+6)dx=-2(x-3)dx\rightarrow \frac{dt}{-2}=(x-3)dx\)

So, the left integral becomes \(\displaystyle \frac{-1}{2}\int \sqrt{1-t^{2}}dt\)

For the right one, let \(\displaystyle u=x-3, \;\ du=dx\)

\(\displaystyle 3\int\sqrt{1-u^{2}}du\)

Now, they can be looked up in a table, trig sub, or what-not. It has to do with arcsin.

Remember to resub.

For #4: Yes, partial fractions is a good idea.

\(\displaystyle \frac{A}{x}+\frac{Bx+C}{x^{2}+1}=3x^{2}+4x+4\)


There is usually more than one way to tackle these integrals. These are just some of the methods that could be used.

 

[Stereotypical]

Wow, nice work! I'm amazed!
But.. about #2 & #4..
I 'm still confused with partial fraction, and especially with the numerators.

For #2: partial fractions.

\(\displaystyle \frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{(x+1)^{2}}=x^{3}\)

For #4: Yes, partial fractions is a good idea.

\(\displaystyle \frac{A}{x}+\frac{Bx+C}{x^{2}+1}=3x^{2}+4x+4\)

What is the criteria of putting "x" in the numerator while using partial fractions?

 

[galactus]

Because of the quadratic term in the denominator.

 

[Deleted member 4993]

Wow, nice work! I'm amazed!
But.. about #2 & #4..
I 'm still confused with partial fraction, and especially with the numerators.




What is the criteria of putting "x" in the numerator while using partial fractions?

That is one of the standard methods when you have polynomial in the denominator.

Study your textbook and class-notes. There is no substitute for it.