Question

Integrate:

Answer 1

\[\int\limits_{}^{}\frac{ 5-\sqrt{x} }{ 5+\sqrt{x} }\]

Answer 2

dx

Answer 3

yeah, sorry, dx

Answer 4

Well, here's as far as I've gotten...

Answer 5

substitute sqrt(x) for u

Answer 6

I end up with \[2\int\limits_{}^{}\frac{ 5u-u ^{2} }{ u+5 }du\]

Answer 7

Then I do long division.

Answer 8

I'd first rationalise the denominator of the integrand: \[\frac{5-\sqrt{x}}{5+\sqrt{x}}\frac{5-\sqrt{x}}{5-\sqrt{x}}=\frac{25-10\sqrt{x}+x}{25-x}\]which then splits our integral to three distinct ones :\[25\int_{}{}\frac{dx}{25-x}\](substitute and you''ll get a natural log ),\[-10\int_{}{}\frac{\sqrt{x}}{25-x}dx\]( suggestions ? ),\[\int_{}{}\frac{x}{25-x}dx\](integration by parts).

Answer 9

\[2\int\limits_{}^{}10-u-\frac{ 50 }{ u+5 }du\]

Answer 10

I tried what you did first, then got stuck with the second part of the integral.

Answer 11

So I got \[20\sqrt{x}-x-100\ln \left| \sqrt{x}+5 \right|+C\]

Answer 12

But it looks wrong to me...

Answer 13

One way to check is to take the derivative of your answer ..... :-)

Answer 14

Hmm... ok. I'll try that.

Answer 15

Man, trying to differentiate this and place it back in that form is as hard as integrating it. Haha.

Answer 16

Yeah, I can't make it match. I must have done something wrong.

Answer 17

I'll check you integration ....

Answer 18

OK\[u=\sqrt{x}=x^{1/2}\]gives\[\frac{du}{dx}=(1/2)x^{-1/2}=\frac{1}{2\sqrt{x}}\]thus\[dx=2\sqrt{x}du=2u\times du\]and\[I= \int_{}{}\frac{5-\sqrt{x}}{5+\sqrt{x}}dx=2\int_{}{}\frac{(5-u)u}{5+u}du\]now\[\frac{-u^2 +5u}{u+5}=\frac{-u^2-5u+5u+5u}{u+5}\]\[=\frac{-u(u+5)+10u}{u+5}=-u + \frac{10u}{u+5}\]\[=-u +\frac{10u+50-50}{u+5}=-u+\frac{10(u+5)-50}{u+5}\]\[=-u + 10-\frac{50}{u+5}\]so\[I=2\int_{}{}[-u + 10-\frac{50}{u+5}]du\]\[=2\frac{-u^{2}}{2}+20u-100ln|u+5|+C\]\[=-x+20\sqrt{x}-100ln|5+\sqrt{x}|+C\]well, wise men agree and fools never differ ! :-)

Answer 19

:) So, does that mean this is the correct answer?

Answer 20

Yeah, I'd pay it. :-)