Question

Integration by substitution help. *question below*

Answer 1

Can I have some help on this question, please?

Answer 2

@jim_thompson5910 @zepdrix @ikram002p

They can help.

Answer 3

Oh you again! :x
the panda lady!

Answer 4

\[\Large\rm \int\limits \frac{1}{1+\sqrt x}(dx)\]
Hmm let's try using the substitution that they suggest, and see what happens.

Answer 5

\[\Large\rm u=1+\sqrt x\]\[\Large\rm du=\frac{1}{2\sqrt x}dx\qquad\to\qquad 2\sqrt x~du=dx\]

Answer 6

In the differential equation.. thing... we need that sqrt(x) in terms of u.
So we'll go back to our original substitution and solve for sqrt(x).

Answer 7

\[\Large\rm u=1+\sqrt x\qquad\to\qquad u-1=\sqrt x\]

Answer 8

\[\Large\rm 2\sqrt x~du=dx\qquad\to\qquad 2(u-1)du=dx\]

Answer 9

Ya? How bout that? :O
That's pretty fancy, huh?

Answer 10

@zepdrix haa, let me process it ^_^

Answer 11

I get that bit :D

Answer 12

So then plug in your goodies,\[\Large\rm \int\limits\limits \frac{1}{1+\sqrt x}[dx]\quad=\quad \int\limits \frac{1}{u}\left[2(u-1)du\right]\]

Answer 13

And it shouldn't be too bad from there :O
What do you think?
Remember how to split up fractions and stuff?

Answer 14

Hmm, split up fractions? :o

Answer 15

I'm lost :s

Answer 16

\[\Large\rm =2\int\limits \frac{u-1}{u}du\]