Question

Calculate the integral

∫1/((1+x)√(2+x))dx

by using the change of variables u=√(x+2)

Answer 1

\[∫1/((1+x)√(2+x))dx\] would then become \[\int\limits_{}^{}\frac{ 1 }{ (1+x)u }dx\]

Answer 2

Can you re-write dx and 1+x in terms of u=Sqrt(2+x)?

Answer 3

No, I don´t know how. Can you help me continue?

Answer 4

\[\large u=\sqrt{2+x}\] \[\large u^2 = 2+x \] \[\large u^2 -1= 1+x \]

Answer 5

ahaa, looks familiar. But how is this useful?

Answer 6

\[\large u=\sqrt{2+x}\]differentiate\[\large du = \frac{ 1 }{ 2 \sqrt {2+x} }dx\]

Answer 7

Original integral was\[\large ∫\frac{ 1 }{ (1+x) }* \frac{1 }{\sqrt{2+x}} dx \]now sub in the above\[\large ∫\frac{ 1 }{ (u^2-1) }* 2du\]Now you can factor and use partial fractions.

Answer 8

please continue if you´ve got the time....

Answer 9

Factor, use partial fractions\[\large \frac{ 1 }{(u+1)(u-1)} = \frac{ A }{ u+1 }+\frac{ B }{ u-1 } \]mult. both sides by the common denom (u+1)(u-1)
\[\large 1 = A(u-1)+B (u+1)\]plug in u=1 to find B=1/2

plug in u=-1 to find A=-1/2


So...\[\large 2∫\frac{ 1 }{ u^2-1 }du = 2\int\limits \left(\frac{ 1/2 }{ u+1 }-\frac{ 1/2 }{ u-1 } \right) du\]
\[\large \int\limits \frac{ 1 }{ u+1 }-\frac{ 1 }{ u-1 } du\]now it should be pretty easy to finish, don't forget to sub back in the expression for u afterwards.

Answer 10

I got the answer ln|u+1|-ln|u-1|+C. Am I near some kind of solution?

Answer 11

Yes.
Now undo your substitution.

Answer 12

What do you mean by undo? Can you show me how?

Answer 13

I think Agent has the signs mixed up from the partial fraction decomposition. We should've gotten,\[\large\rm \int\limits\frac{1}{u-1}-\frac{1}{u+1}du\]Leading to,\[\large\rm =\ln|\color{orangered}{u}-1|-\ln|\color{orangered}{u}+1|+c\]Remember, your original integral was given in terms of x, not u, so your answer should also be in terms of x.\[\large\rm =\ln|\color{orangered}{\sqrt{2+x}}-1|-\ln|\color{orangered}{\sqrt{2+x}}+1|+c\]

Answer 14

ok, thanks! :)