Integration by Parts: Evaluate ∫ 1 ÷ (16 + x2) dx

[Christina-]

Homework Statement

Use integration by parts to evaluate the integral:
∫ 1 ÷ (16 + x²) dx

Homework Equations

∫ u dv = uv - ∫ v u' du

The Attempt at a Solution

That's the problem, I don't know how to start. How would I divide up 1/(16 + x²) into two? So there would be a value for u and v'.

Maybe this isn't so much a question of how do you solve the integral, but how do you split the above polynomial. There's also tan in the answer, but I'm not sure how to get to that.
Any help to point me in the right direction would be greatly appreciated.

 

[quasar987]

Are you sure the problem asks you to do this integral by parts?

The derivative of Arctg(x) is 1/(1 + x²) so simply factoring 1/16 then performing a change of variable y=x/4 gives the result.

 

[dextercioby]

It can be done by parts, but it's also interesting to do it by simple fraction expansion.

 

[Schrodinger's Dog]

uneditable. See below.

 

[cristo]

If I can do it either way then it must be easy

Spoiler:-

$$\int \frac{1}{16+x^2} => \frac {1}{16}+\frac{1}{x^2} =>\int \frac {1}{16}+x^{-2}$$

This is incorrect: $$\frac{1}{16+x^2}\neq\frac{1}{16}+\frac{1}{x^2}$$

substitute du.=arctan g(x) integrate using the sum rule and then simplify.

$$= \frac{1}{4}. tan^{-1}. \frac {4}{x}+C$$

There is an error here, it should read: $$= \frac{1}{4}. tan^{-1}. \frac {x}{4}+C$$

 

[Schrodinger's Dog]

This is incorrect: [tex]\frac{1}{16+x^3}\neq\frac{1}{16}+\frac{1}{x^2}

There is an error here, it should read: $$= \frac{1}{4}. tan^{-1}. \frac {x}{4}+C$$

Tell me about it. I keep trying to correct this but am getting nowhere fast. Thanks for the tex. I'll alter it so it reads correctly, my bad. I realized my mistake and have spent the last ten minutes trying to correct it bear with me.

I can't edit it any further but you should end up with

$$= \frac{1}{4}. tan^{-1}. \frac {x}{4}+C$$

That was an hour well spent certainly learned a lot about editting :rofl:

It can be done by parts, but it's also interesting to do it by simple fraction expansion.

If I can do it either way then it must be easy

$$\int\frac{1}{(16+x^2)} = \frac{1}{16(1+1/16x^2)}$$

It's pretty straight forward from here.

Spoiler.

$$\int \frac {1}{16}(1+x^2) where = \frac{1}{1+x^2}=tan^{-1}g(x)=>$$$$\frac{1}{4}. tan^{-1}. \frac {x}{4}+C$$

Sorry