CALCULUS - INTEGRATION OF INVERSE TRIGONOMETRIC FUNCTIONS

Question

henryarias14 · Answer

Find the indefinite Integral

$$\int\limits_{}^{} \frac{ x-3 }{ x^2+1 } dx$$

henryarias14 · Answer

the book gives me an answer with natural logs and i dont know how they got there. please help through out the steps

henryarias14 · Answer

maybe using log division first?

zepdrix · Answer

Answer key didn't include an arctangent..?

henryarias14 · Answer

yeah it did

henryarias14 · Answer

natural log minus and arctan

henryarias14 · Answer

to be honest log division is the only thing i see but im not sure

zepdrix · Answer

Here is one approach,$$\large\rm \int\limits\frac{x-3}{x^2+1}~dx\quad=\quad \int\limits\frac{x}{x^2+1}~dx\int\limits\frac{-3}{x^2+1}~dx$$you can split this into two separate fractions.

First one is simple u-substitution.
Second one is arctangent.

zepdrix · Answer

woops lemme fix that >.<

zepdrix · Answer

$$\large\rm \int\limits\limits\frac{x-3}{x^2+1}~dx\quad=\quad \int\limits\limits\frac{x}{x^2+1}~dx+\int\limits\limits\frac{-3}{x^2+1}~dx$$

henryarias14 · Answer

but i dont know because the denominator is x^2 and okay lol

zepdrix · Answer

Are you ok with the fraction business?

From there you would make the substitution $\large\rm u=x^2+1$
Your du will be something close to the numerator.

henryarias14 · Answer

yeah im okay with fractions and oh i see let me try it

henryarias14 · Answer

well du is 2x so what would be the fraction?

henryarias14 · Answer

du-3 all over 2?

henryarias14 · Answer

i mean du -6 all over 2?

henryarias14 · Answer

$$\frac{ du-6 }{ 2 }$$

zepdrix · Answer

Let's move the dx into the numerator just so it's a little more clear what is going on,$$\large\rm \int\limits\limits\limits\frac{x-3}{x^2+1}~dx\quad=\quad \int\limits\limits\limits\frac{x~dx}{x^2+1}+\int\limits\limits\limits\frac{-3}{x^2+1}~dx$$

So you made the substitution $\large\rm u=x^2+1$
And you found the differential to be $\large\rm du=2x~dx$
Dividing by 2 gives us exactly what we have in the numerator, yes?
$\large\rm \frac12du=x~dx$

zepdrix · Answer

$$\large\rm \color{royalblue}{\int\limits\limits\limits\limits\frac{x~dx}{x^2+1}}+\int\limits\limits\limits\limits\frac{-3}{x^2+1}~dx$$And just in case there is any confusion,
we're only applying this substitution to the first integral (in blue here).

henryarias14 · Answer

ohhhh you split them into two now i get it

zepdrix · Answer

Remember that your answer key involves a natural log and an arctangent.

That's why we want to deal with them as two separate integrals,
because we expect very different results from each one.

henryarias14 · Answer

so $$\frac{ 1 }{ 2 }\int\limits_{}^{} \frac{ du }{ u }$$

henryarias14 · Answer

something like that?

zepdrix · Answer

$$\large\rm \color{royalblue}{\frac12\int\limits\frac{du}{u}}+\int\limits\limits\limits\limits\limits\frac{-3}{x^2+1}~dx$$Mmm yup, looks good!

henryarias14 · Answer

now i have a question for you. if you were taking an exam and didnt know what the answer would be ahead of time what would be your first impression of this problem, what would be your first course of action

henryarias14 · Answer

because for this one i told you it would be a natural log - arctan but what if you didnt know that

zepdrix · Answer

My first instinct is to look at the `powers of x`.

I see a square x in the denominator,
and a linear x in the numerator.

So a light bulb starts to flicker, and I think "Hmm there might be a u-substitution in here somewhere..."

So you try $\large\rm u=x^2+1$
But then you find out that $\large\rm du\ne (x-3)dx$
It's not close to the numerator unfortunately.

So then maybe we need to apply some algebra before a u-substitution will work.

zepdrix · Answer

So that was my thought process in breaking it into two separate fractions.
I really really want a u-substitution to work.

And it turns out that the x alone gives us what we want.

henryarias14 · Answer

i see. now, i thought you couldn't split them up into two fractions because the numerator is a summation. do you know the exact rule for that? because that part of my algebra is kind of shaky

henryarias14 · Answer

in other words, when can you and when can you not split them up into two separate fractions?

zepdrix · Answer

|dw:1481764211742:dw|Quick fraction refresher here.
I like to call this the jellybean method ... or something :p

You just draw these big ugly jelly beans around each term in the numerator and the WHOLE denominator. You can split numerators, but not denominators.