# Thread: integral , basic ..

1. ## integral , basic ..

Use this formula only $\int x^n ~dx = \frac{ x^{n+1}}{n+1} + c$

Find the integral $\int \frac{dx}{\sqrt{ x + \sqrt{ x^2 + 1}}}$

Of course , little Algebra is allowed . Try to answer this without applying substitution , integration by parts , etc

2. Edit

3. What is the solution
i lost 2 hours from my life
and BTW, you should said that n dont equal -1

4. ## intermediate step

I'm considering squaring.

5. It seems impossible to solve without using substitution

6. Below is the solution to this problem ,

Spoiler:

To avoid appearing the imaginary number $i$ in the solution ,

Consider

$\int \frac{dx}{\sqrt{ x + \sqrt{ x^2 - a^2 }}}$

$= \int \frac{dx}{\sqrt{ x + \sqrt{ x^2 + 1}}} \cdot \frac{ \sqrt{ x - \sqrt{x^2 - a^2 }} }{\sqrt{ x - \sqrt{x^2 - a^2 }}}$

$= \int \frac{ \sqrt{ x - \sqrt{x^2 - a^2 }} }{ \sqrt{ x^2 - (x^2 - a^2)}}~dx$

$= \frac{1}{a} \int \sqrt{ x - \sqrt{x^2 - a^2 }} ~dx$

Now , the integral looks much more simplier but that's not enough , we have to do something special in the following steps :

$= \frac{1}{a} \int \sqrt{\frac{2( x - \sqrt{x^2 - a^2})}{2}} ~dx$

$= \frac{1}{a \sqrt{2}} \int \sqrt{2x - 2\sqrt{(x-a)(x+a)}}~dx$

$= \frac{1}{a \sqrt{2}} \int \sqrt{ ( x+a) - 2\sqrt{(x-a)(x+a)} + (x-a)}~dx$

$= \frac{1}{a \sqrt{2}} \int \sqrt{ \sqrt{x+a}^2 - 2\sqrt{(x-a)(x+a)} + \sqrt{x-a}^2}~dx$

$= \frac{1}{a \sqrt{2}} \int \sqrt{ ( \sqrt{x+a} - \sqrt{x-a})^2}~dx$

$= \frac{1}{a \sqrt{2}} \int [\sqrt{x+a} - \sqrt{x-a} ]~dx$

$= \frac{\sqrt{2}}{3a} \left[ \sqrt{x+a}^3 - \sqrt{x-a}^3 \right ] + C$

$= \frac{\sqrt{2}}{3a} ( \sqrt{x+a} - \sqrt{x-a})( (x+a) + \sqrt{x^2 - a^2} + (x-a) ) + C$

$= \frac{2}{3a} \sqrt{ x - \sqrt{x^2 - a^2}} ( 2x + \sqrt{ x^2 - a^2}) + C$

$= \frac{2}{3} \frac{1}{\sqrt{x+ \sqrt{x^2 - a^2}}} ( 2x + \sqrt{x^2 - a^2} ) + C$

Sub. $-a^2 = +1$

the integral

$= \frac{2}{3} \frac{2x + \sqrt{x^2 +1} }{\sqrt{x+ \sqrt{x^2 +1}}} + C$

7. Originally Posted by TWiX
What is the solution
i lost 2 hours from my life
and BTW, you should said that n dont equal -1

I just forgot to add this point but actually it is also true when $n = -1$

Although , from the formula , it gives us something like this :

$\frac{ x^0}{0} + C$ or $\lim_{a\to 0} \frac{x^a}{a} + C$

the numerator tends to $1$ while the denominator tends to zero .

,if we rewrite the constant $C$ as $-\frac{1}{a} + C'$ , and make $a$ very small $\to 0$

so we have

$\int \frac{dx}{x} = \lim_{a\to 0} \frac{x^a -1 }{a} + C'$

Then use L rule to evaluate the limit , we will find that

$I = \ln(x) + C'$