Question

Please help me get started on the following integral (click to see).

Answer 1

\[\int\limits_{}^{}\frac{ 1 }{ \sqrt{x}(x+1) }dx\]

Answer 2

Okay, first thing I would suggest is distributing the radical x on the bottom there.

Answer 3

u = sqrt(x) perhaps.

Answer 4

u = sqrt(x) would give you a problem when you find du. But u-substitution would be the best method of approach.

Answer 5

\[\int\limits_{}^{}\frac{ 1 }{ x ^{3/2}+\sqrt{x} }dx\]

Answer 6

u=x
?

Answer 7

But du = dx/2sqrt(x) which you have

Answer 8

Hmm, I see where you're going, but if you just set u = 1/sqrt(x) to begin with your substituted integral will be simpler.

Answer 9

substituting u=x wouldn't end up changing the problem at all babyslap, I know it's tempting to say that when you're at a loss but remember you're using substitution to try and simplify the problem, not just replace x with a different variable.

Answer 10

oh right

Answer 11

u = sqrt(x)
du = dx/2sqrt(x)
Integral becomes 2du/(u^2+1) which is easy

Answer 12

\[\int\limits_{}^{}x ^{-3/2}+x ^{-1/2}dx\]

Answer 13

No sub necessary then right?

Answer 14

integral becomes -2/u du if you sub 1/sqrt(x) though.

Answer 15

Fractions don't work like that baby 1/(2+3) =/= 1/2 + 1/3

Answer 16

Just try the substitution u=sqrt(x). It works

Answer 17

ok

Answer 18

could i also solve by integrating by partial fractions?

Answer 19

yes

Answer 20

Yes but it's more trouble than it's worth. If you can learn to recognize these kinds of substitutions you can just solve the integral in your head.