Definite integral (exact value): Help solving this definite integral

Question

Here's the integral:

$$\int_{1}^{5}{\frac{\sqrt{x}}{\sqrt{6-x}+\sqrt{x}}dx}$$

tried rationalising the function:

$$\frac{\sqrt{x\left( 6-x \right)}-x}{6-2x}$$

which doesn't get me anywhere with substitution or integration by parts

I would appreciate if someone were to help me through hints/techniques (besides numerical methods).

See also: http://math.stackexchange.com/questions/578957/definite-integral-int-24-frac-sqrt-log9-x-sqrt-log9-x-sqrt-log3 — lab bhattacharjee, Sep 22 '15 at 08:33

juantheron · Accepted Answer · 2015-09-22T08:31:32.843

2

Let $$\displaystyle I = \int_{1}^{5}\frac{\sqrt{x}}{\sqrt{6-x}+\sqrt{x}}dx.................(1)$$

Now Using $$\displaystyle \bullet \int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx$$

So Replace $x\rightarrow (1+5-x) = (6-x)$ in equation $(1)$

So we get $$\displaystyle I = \int_{1}^{5}\frac{\sqrt{6-x}}{\sqrt{x}+\sqrt{6-x}}dx..........(2)$$

Now add above two equation, we get

$$\displaystyle 2I = \int_{1}^{5}\frac{\sqrt{x}+\sqrt{6-x}}{\sqrt{6-x}+\sqrt{x}}dx = \int_{1}^{5}1dx =\left[x\right]_{1}^{5}= 4$$

So we get $$\displaystyle \displaystyle I = \int_{1}^{5}\frac{\sqrt{x}}{\sqrt{6-x}+\sqrt{x}}dx= 2$$

edited Sep 22 '15 at 08:31

answered Sep 22 '15 at 08:28

juantheron

The limit should be $a$ to $b$ not $1$ to $b$ – lab bhattacharjee Sep 22 '15 at 08:30
so you took $u=a+b-x$ and the limits should be b to a – inspd Sep 22 '15 at 08:31
is it a to b or – inspd Sep 22 '15 at 08:32
Thanks lab bhattacharjee I have edited it. Yes inspad it is from $a$ to $b$ – juantheron Sep 22 '15 at 08:32
why, when $x=a$ $u=b$ – inspd Sep 22 '15 at 08:33
oh forgot the $du/dx$ nvm – inspd Sep 22 '15 at 08:33
Now how would I realise such a thing under exam conditions.. – inspd Sep 22 '15 at 08:34
You should read properties of definite Integrals and examples related to such properties. Then you will able to solve these type of questions. – juantheron Sep 22 '15 at 08:35
do you have a good link on these? i'm googling it now anyway – inspd Sep 22 '15 at 08:36

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