Question

help finding the indefinite integral

Answer 1

\[\int\limits \frac{ x }{ \sqrt{x+3} } dx\]

Answer 2

You want to \(u\) sub with \(u=x+3\) I think.

Answer 3

u=x+3, du/dx = 1, dx=du
\[\int\limits \frac{ x }{ \sqrt{x+3}} = \int\limits \frac{ x }{ \sqrt{u} }du\]

Answer 4

you still need to write x in terms of u

Answer 5

if u=x+3
then x=u-3

Answer 6

you can then write your integrand as two fractions

Answer 7

\[\int\limits \frac{u-3}{u^\frac{1}{2}} du=\int\limits(\frac{u}{u^{\frac{1}{2}}}-\frac{3}{u^\frac{1}{2}} ) du\]
use law of exponents
then user the antiderivative power rule

Answer 8

\[\int\limits u^n du=\frac{u^{n+1}}{n+1} +C \]
this rule works for any n except n equals -1

Answer 9

\[=(\frac{ u ^{2} }{ 2 }+3u ) \div \frac{ 2u ^{3/2} }{ 3 }\]

Answer 10

is that right?

Answer 11

how did you get that?

Answer 12

It doesn't look right :p

Answer 13

did you ever use the law of exponents I mentioned ?

Answer 14

yes

Answer 15

so you used u^a/u^b=u^(a-b)
and 1/u^(-b)=u^b

Answer 16

\[\int\limits\limits \frac{u-3}{u^\frac{1}{2}} du=\int\limits\limits(\frac{u}{u^{\frac{1}{2}}}-\frac{3}{u^\frac{1}{2}} ) du \\ =\int\limits (u^{1-\frac{1}{2}}-3u^{\frac{-1}{2}} ) du \\ \int\limits (u^\frac{1}{2}-3u^{\frac{-1}{2}} ) du\]

Answer 17

and see you can write as two integrals
and use that one power for integrals thingy

Answer 18

power rule*

Answer 19

\[\frac{ 2u ^{3/2} }{ 3} - 3 \times 2\sqrt{u}\]

Answer 20

+c
yes

and you can write 3*2 as 6

Answer 21

and then recall we started in terms of x
so we need to finish in terms of x

Answer 22

u=x+3
so you replace u with x+3

Answer 23

got it, thank you!