Question

Okay, another integral.
(problem inside) Hints. don't do my work:) I'll see what I can do and what I can't.

Answer 1

\[\Large \int\limits_{ }^{ } \frac{1}{\sqrt{\theta-1}+\sqrt{(\theta-1)^3}}d\theta\]

Answer 2

Well, firstly, I would substitute,
\[\LARGE w=\theta-1~~~~~dw=d \theta\](tired of U s)

Answer 3

\[\Large \int\limits_{ }^{ } \frac{1}{\sqrt{w}+\sqrt{w^3}}dw\]

Answer 4

factor, and use substitutions. check the pattern.

Answer 5

Factor?

Answer 6

Can we come back to the factoring later, ones I finish, if I be able to?

Answer 7

the denominator, that was just a random guess...II haven't even started solving it myself :)

Answer 8

and then what, partial fraction?

Answer 9

I would probably pick sqrt(theta-1)=u instead.

Answer 10

Ohh, okay, I'll try this.

Answer 11

\[\Large \int\limits_{ }^{ } \frac{1}{(\sqrt{\theta -1 }+\sqrt{\theta -1 }(\theta -1 )^2}d\theta\]

but there will be a difficulty with du/d(theta) part though

Answer 12

Oops, the ^2 shouldn't be there.

Answer 13

Let's first do the w sub. we can always come back if we want, can't we?

Answer 14

\[\Large \int\limits\limits_{ }^{ } \frac{1}{(\sqrt{w}+w\sqrt{w})}dw\]

Answer 15

it is just looks better.

Answer 16

Well there's more than one way to do this problem, I think after the substitution I've done it becomes much nicer and obvious.

Answer 17

I'll try partial fractions. I am new to it, but I saw a couple examples from the book.
We didn't learn this method in class yet.

Answer 18

I would probably rewrite your square roots as w^1/2 instead. I did it without partial fractions.

Answer 19

\[\Large \int\limits\limits_{ }^{ } \frac{A}{(\sqrt{w}}+\frac{B}{w}dw\]\[\Large A(w)+B(\sqrt{w})=1\] \[\Large A+B=1\]\[\Large 4A+2B=1\]

Answer 20

I plugged x=1, and then x=4.

Answer 21

\[\Large B=1-A\]

Answer 22

I don't think you can do partial fractions like that.

Answer 23

yes I can, I am fairly, sure that the A(w)+B(sqrt w) =1 is true for all values.

Answer 24

MY dad is calling, I got to go. Sorry for disappointing y'all.

Answer 25

I'll close it for now.

Answer 26

I'll write up my answer, and you can see what I did later.

Answer 27

\[\Large \int\frac{1}{(\sqrt{w}+w\sqrt{w})}dw=\Large \int\frac{1}{\sqrt{w}}\frac{1}{(1+(\sqrt{w})^2)}dw\]

Answer 28

\[=\Large \int\frac{1}{2\sqrt{w}}\frac{2}{(1+(\sqrt{w})^2)}dw\]

Answer 29

\[\Large u = (\theta -1)^\frac{1}{2} \\ \Large 2du = \frac{1}{(\theta - 1 )^\frac{1}{2}}d \theta\] now factor the integral like this: \[\Large \int\limits \frac{1}{1+(\theta-1)} \frac{d \theta}{ (\theta-1)^\frac{1}{2}}\] Now be careful plugging it in and pay attention to what I've done: \[\Large \int\limits \frac{1}{1+u^2} 2 du\]
You should recognize this integral. =P