1. ## integrate by substitution

how do you integrate 1/ ( 6+x+ x^2 )^(1/2)

my working:

let 6+x+ x^2 = (x+2)^2 (z^2)

then dx = -10z / ( z^2 +1) ^(2) dz

substituting that into the question, i got

integrate -2 / (z^2+1) and hence the answer was -2 arctan ( ((3-x)/ (x+2))^(1/2))

however the answer should have been arcsin ( (2x-1)/5)).

is there something wrong with my substitution?

2. Have you considered Completing the Square and then a Trigonometric substitution?

3. Originally Posted by alexandrabel90
how do you integrate 1/ ( 6+x+ x^2 )^(1/2)

my working:

let 6+x+ x^2 = (x+2)^2 (z^2)

then dx = -10z / ( z^2 +1) ^(2) dz

substituting that into the question, i got

integrate -2 / (z^2+1) and hence the answer was -2 arctan ( ((3-x)/ (x+2))^(1/2))

however the answer should have been arcsin ( (2x-1)/5)).

is there something wrong with my substitution?
$\int{\frac{1}{\sqrt{x^2 + x + 6}}\,dx} = \int{\frac{1}{\sqrt{x^2 + x + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 + 6}}\,dx}$

$= \int{\frac{1}{\sqrt{\left(x + \frac{1}{2}\right)^2 + \frac{23}{4}}}\,dx}$.

Now use the substitution $\frac{\sqrt{23}}{2}\tan{\theta} = x + \frac{1}{2}$

So $\frac{dx}{d\theta}= \frac{\sqrt{23}}{2}\sec^2{\theta}$

$dx = \frac{\sqrt{23}}{2}\sec^2{\theta}\,d\theta$.

So the integral becomes

$\int{\frac{1}{\sqrt{\left(\frac{\sqrt{23}}{2}\tan{ \theta}\right)^2 + \frac{23}{4}}}\,\frac{\sqrt{23}}{2}\sec^2{\theta}\ ,d\theta}$

$= \frac{\sqrt{23}}{2}\int{\frac{1}{\sqrt{\frac{\sqrt {23}}{2}(\tan^2{\theta} + 1)}}\,\sec^2{\theta}\,d\theta}$

$= \frac{\sqrt{23}}{2}\int{\frac{1}{\sqrt{\frac{\sqrt {23}}{2}(\sec^2{\theta})}}\,\sec^2{\theta}\,d\thet a}$

$= \frac{\sqrt{23}}{2}\int{\frac{1}{\sqrt{\frac{\sqrt {23}}{2}}\sec{\theta}}\,\sec^2{\theta}\,d\theta}$

$= \frac{\sqrt{23}}{2}\int{\frac{1}{\sqrt{\frac{\sqrt {23}}{2}}\sec{\theta}}\,\sec^2{\theta}\,d\theta}$

$= \frac{\sqrt{23}}{2}\,\sqrt{\frac{2}{\sqrt{23}}}\in t{\sec{\theta}\,d\theta}$

$= \frac{\sqrt{23}}{2}\,\sqrt{\frac{2\sqrt{23}}{23}}\ int{\sec{\theta}\,d\theta}$

$= \frac{\sqrt{2\sqrt{23}}}{2}\int{\sec{\theta}\,d\th eta}$

$= \frac{\sqrt{2\sqrt{23}}}{2}\ln{|\sec{\theta} + \tan{\theta}|}$

Now get this function back in terms of $x$.

Remember that

$\frac{\sqrt{23}}{2}\tan{\theta} = x + \frac{1}{2}$

and $\sec{\theta} = \sqrt{1 + \tan^2{\theta}}$.

P.S. I do not agree with the answer given. If they want the integral to contain arcsin, the integrand should have been of the form $\frac{1}{\sqrt{a^2 - x^2}}$, not $\frac{1}{\sqrt{a^2 + x^2}}$ as yours was.

Are you sure you copied the question down correctly?

4. sorry the question should have been
integrate 1/ ( 6+x- x^2 )^(1/2) instead.

5. Originally Posted by alexandrabel90
how do you integrate 1/ ( 6+x+ x^2 )^(1/2)

my working:

let 6+x+ x^2 = (x+2)^2 (z^2)

then dx = -10z / ( z^2 +1) ^(2) dz

substituting that into the question, i got

integrate -2 / (z^2+1) and hence the answer was -2 arctan ( ((3-x)/ (x+2))^(1/2))

however the answer should have been arcsin ( (2x-1)/5)).

is there something wrong with my substitution?
Upon doing some further research, this question is much easier using a hyperbolic substitution, rather than a trigonometric.

$\int{\frac{1}{\sqrt{\left(x + \frac{1}{2}\right)^2 + \frac{23}{4}}}\,dx}$.

So you let $x + \frac{1}{2} = \frac{\sqrt{23}}{2}\sinh{t}$.

Therefore $\frac{dx}{dt} = \frac{\sqrt{23}}{2}\cosh{t}$, and so $dx = \frac{\sqrt{23}}{2}\cosh{t}\,dt$.

So the integral becomes

$\int{\frac{1}{\sqrt{\left(\frac{\sqrt{23}}{2}\sinh {t}\right)^2 + \frac{23}{4}}}\,\frac{\sqrt{23}}{2}\cosh{t}\,dt}$

$= \frac{\sqrt{23}}{2}\int{\frac{1}{\sqrt{\frac{23}{4 }(\sinh^2{t} + 1)}}\,\cosh{t}\,dt}$

$= \frac{\sqrt{23}}{2}\int{\frac{1}{\frac{\sqrt{23}}{ 2}\sqrt{\cosh^2{t}}}\,\cosh{t}\,dt}$

$= \frac{\sqrt{23}}{2}\,\frac{2}{\sqrt{23}}\int{\frac {1}{\cosh{t}}\,\cosh{t}\,dt}$

$= \int{1\,dt}$

$= t + C$.

Now since $x + \frac{1}{2} = \frac{\sqrt{23}}{2}\sinh{t}$

$\sinh{t} = \frac{2}{\sqrt{23}}\left(x + \frac{1}{2}\right)$

$\sinh{t} = \frac{2x + 1}{\sqrt{23}}$

$t = \sinh^{-1}{\left(\frac{2x + 1}{\sqrt{23}}\right)}$.

Therefore the integral is

$\sinh^{-1}{\left(\frac{2x + 1}{\sqrt{23}}\right)} + C$.

6. Originally Posted by alexandrabel90
sorry the question should have been
integrate 1/ ( 6+x- x^2 )^(1/2) instead.
Right.

$\int{\frac{1}{\sqrt{6 + x - x^2}}\,dx} = \int{\frac{1}{\sqrt{-(x^2 - x - 6)}}\,dx}$

$= \int{\frac{1}{\sqrt{-\left[x^2 - x + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 - 6\right]}}\,dx}$

$= \int{\frac{1}{\sqrt{-\left[\left(x - \frac{1}{2}\right)^2 - \frac{25}{4}\right]}}\,dx}$

$= \int{\frac{1}{\sqrt{\frac{25}{4} - \left(x - \frac{1}{2}\right)^2}}\,dx}$.

Now use the substitution $x - \frac{1}{2} = \frac{5}{2}\sin{\theta}$.

7. instead of using completing the square, can i solve the question if i factorise
6+x-x^2 = (3-x)(x+2) ?

8. Originally Posted by alexandrabel90
instead of using completing the square, can i solve the question if i factorise
6+x-x^2 = (3-x)(x+2) ?
No. The square root in the denominator prevents you from being able to factorise and use partial fractions.

Have a go using the trigonometric subsitution.