Indefinite Integration by exchange of variables

[Asphyxiated]

Homework Statement

\int \frac {1}{\sqrt{x}(1+\sqrt{x})} dx

The Attempt at a Solution

So let

u = 1 + \sqrt{x}

then

du = \frac {1}{2}x^{-1/2} dx

So dx should be this:

dx = 2x^{1/2} du

right?

So now the Problem looks something like this:

\int \frac {2\sqrt{x}}{\sqrt{x}(u)} du

\int \frac {\sqrt{x}}{(u)} du

I am stuck here, I am just not sure where to go from here, the answer is suppose to be:

2ln(1+\sqrt{x}) + C

but I really don't know how to get there.

thanks!

 

[zachzach]

Homework Statement

\int \frac {1}{\sqrt{x}(1+\sqrt{x})} dx

The Attempt at a Solution

So let

u = 1 + \sqrt{x}

then

du = \frac {1}{2}x^{-1/2} dx

So dx should be this:

dx = 2x^{1/2} du

right?

So now the Problem looks something like this:

\int \frac {2\sqrt{x}}{\sqrt{x}(u)} du

\int \frac {\sqrt{x}}{(u)} du

I am stuck here, I am just not sure where to go from here, the answer is suppose to be:

2ln(1+\sqrt{x}) + C

but I really don't know how to get there.

thanks!

\int \frac {2\sqrt{x}}{u\sqrt{x}} du = \int \frac {\sqrt{x}}{u} du ??

\int \frac {2\sqrt{x}}{u\sqrt{x}} du = \int \frac {2}{u} du

 

[Asphyxiated]

haha right... so then the proper way to get to the answer is this then:

\int \frac {2\sqrt{x}}{u\sqrt{x}} du = \int \frac {2}{u} du

\int 2 * \frac {1}{u} du

then take an antiderivative:

2*ln(u)

2ln(1+\sqrt{x}) +C

is that right?

 

[zachzach]

haha right... so then the proper way to get to the answer is this then:

\int \frac {2\sqrt{x}}{u\sqrt{x}} du = \int \frac {2}{u} du

\int 2 * \frac {1}{u} du

then take an antiderivative:

2*ln(u)

2ln(1+\sqrt{x}) +C

is that right?

Looks right to me.

 

[Asphyxiated]

ok can you help me on a problem of the same type?

\int \frac {3t+4}{5-t}dt

Let:

u = 5 -t

du = -1 dt

dt = -1 du

\int -\frac {3t+4}{u} du

so here I am kind of guessing..

\int -(3t+4)* \frac {1}{u} du

\int (-3t-4) * \frac {1}{u}du

\int -3t\frac{1}{u}-4\frac{1}{u}

I don't think I am on the right track here though, the correct answer is

-3t -19ln(5-t) +C

and I don't see how to get there from here, its probably just something I am not seeing again... thanks in advance

 

[zachzach]

ok can you help me on a problem of the same type?

\int \frac {3t+4}{5-t}dt

Let:

u = 5 -t

du = -1 dt

dt = -1 du

\int -\frac {3t+4}{u} du

so here I am kind of guessing..

\int -(3t+4)* \frac {1}{u} du

\int (-3t-4) * \frac {1}{u}du

\int -3t\frac{1}{u}-4\frac{1}{u}

I don't think I am on the right track here though, the correct answer is

-3t -19ln(5-t) +C

and I don't see how to get there from here, its probably just something I am not seeing again... thanks in advance

u = 5 -t so
t = 5 -u
3t + 4 = 3(5-t)+4 = 19 -3u

Try to figure it out from there. Hint: split into 2 integrals. You have basically right idea, you just need to get that t in terms of u somehow so you can actually integrate.

 

[Asphyxiated]

ok so starting from the point of

\int (-19+3u)*\frac{1}{u} du

note that I just applied the - sign throughout, that's why my signs are different than yours

so my next guess would be here:

\int -19\frac{1}{u} + 3u\frac{1}{u} du

so then I am assuming I can like so:

\int -19\frac{1}{u} + 3u\frac{1}{u} du = \int -19\frac{1}{u} du + \int \frac {3u}{u} du

so I can see that the -19ln(5-t) comes out there but the integral of 3 du is 3u right? so 3(5-t) = 15-3t which does give me the -3t for the solution but the +15 ruins the answer... at least I think so...

 

[zachzach]

ok so starting from the point of

\int (-19+3u)*\frac{1}{u} du

note that I just applied the - sign throughout, that's why my signs are different than yours

so my next guess would be here:

\int -19\frac{1}{u} + 3u\frac{1}{u} du

so then I am assuming I can like so:

\int -19\frac{1}{u} + 3u\frac{1}{u} du = \int -19\frac{1}{u} du + \int \frac {3u}{u} du

so I can see that the -19ln(5-t) comes out there but the integral of 3 du is 3u right? so 3(5-t) = 15-3t which does give me the -3t for the solution but the +15 ruins the answer... at least I think so...

I got the same answer as you and that looks right. The answer you got it:

y = 15 - 3t -19ln(5-t) + c_1

Let

c = 15 + c_1 \rightarrow y = -3t - 19ln(5-t) + c

The +15 is just a constant so it does not change the functional form of your answer. That is why you might as well absorb it into the c_1 so your answer looks nicer.

 

[Asphyxiated]

Wow I actually didn't know that, thanks man, I would have been thinking that over forever.