Question

Evaluate the indefinite integral. (Remember to use ln |u| where appropriate.)

1- (x)^(1/2) / 5+ (x)^(1/2)

Answer 1

\[\large \int\limits \frac{1-\sqrt x}{5+\sqrt x}dx\]

This one is a bit tricky. I had to look at wolfram to get a hint :p
We want to make the substitution, \(\large u=\sqrt x\).

Answer 2

\[\large \int\limits \frac{1-u}{5+u}\;\color{orangered}{dx}\]So now we need to figure out something for this dx here.

Answer 3

So taking the derivative of our substitution gives us,\[\large u=\sqrt x \qquad \rightarrow \qquad du=\frac{1}{2\sqrt x}\color{orangered}{dx}\]Remember, we're trying to solve for dx here, so let's multiply both sides by 2sqrtx.\[\large \color{orangered}{dx}=2\sqrt x \;du\]And recall that we said that sqrtx=u, so let's sub that in,\[\large \color{orangered}{dx}=2u\;du\]

Answer 4

So that allows us to rewrite our integral fully in terms of \(\large u\).
\[\large \int\limits\limits \frac{1-u}{5+u}\;\color{orangered}{dx} \qquad \rightarrow \qquad \int\limits\limits \frac{1-u}{5+u}\;\color{orangered}{\left(2u\;du\right)}\]

Answer 5

\[\large 2\int\limits \frac{u-u^2}{5+u}\;du\]

Hmm, the degree of the top is larger than the degree of the bottom.
You'll want to apply `Polynomial Long Division` to simplify this.
Do you remember how to do that? :o

Answer 6

so it becomes 1-u + (5u-5)/(u+5)

Answer 7

Hmm the degree of the top is now `equal` to the degree of the bottom. In which case you would want to apply polynomial long division.

Maybe you just didn't divide far enough <:O I dunno.

I'm coming up with something like this when you divide them,\(\large -u+6\) with a remainder of \(\large -30\)

Answer 8

\[\large 2\int\limits\limits \frac{u-u^2}{5+u}\;du \qquad =\qquad 2\int\limits 6-u-\frac{30}{5+u}du\]

Answer 9

would it be 12(5+squareroot x) + 2(25+10squareootx +x)/2 -10ln(10+squarerootx)

Answer 10

would that be right?

Answer 11

nevermind i got it!

Answer 12

Thank you!