How do you find the integral
#intx^2/((x^2+2)^(3/2))dx# ?

Question

How do you find the integral
 #intx^2/((x^2+2)^(3/2))dx# ?

Emilia Aguilar · Accepted Answer

#=1/2*x/sqrt(2+x^2)+c#, where #c# is a constant

Explanation :

#=intx^2/((x^2+2)^(3/2))dx#

#=intx^2/(x^3(1+2/x^2)^(3/2))dx#

#=int1/(x(1+2/x^2)^(3/2))dx#

Using Integration by Substitution,

let's assume #2/x^2=t#

then #-4/xdx=dt#

#=int-dt/(4(1+t)^(3/2))#

#=-1/4int(1+t)^(-3/2)dt#

#=-1/4*((1+t)^(-1/2))/(-1/2)#

#=1/2*1/sqrt(1+t)+c#, where #c# is a constant

Substituting #t# back,

#=1/2*1/sqrt(1+2/x^2)+c#, where #c# is a constant

#=1/2*x/sqrt(2+x^2)+c#, where #c# is a constant

William Abdalla · Answer

undefined To find the integral ∫x^2/((x^2+2)^(3/2)) dx, you can use the substitution method. Let $ u = x^2 + 2 $. Then, $ du = 2x dx $. Rearranging gives $ x dx = \frac{1}{2} du $. Substituting these into the integral gives:

$$ \int \frac{x^2}{(x^2+2)^{3/2}} dx = \int \frac{1}{2} \frac{1}{u^{3/2}} du $$

Now, integrate with respect to $ u $:

$$ = \frac{1}{2} \int u^{-3/2} du $$

$$ = \frac{1}{2} \cdot \frac{u^{-1/2}}{-1/2} + C $$

$$ = -\frac{1}{u^{1/2}} + C $$

Substitute back $ u = x^2 + 2 $:

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

So, the integral is $ -\frac{1}{\sqrt{x^2 + 2}} + C $, where $ C $ is the constant of integration.

William Abdalla · Answer

undefined To find the integral $ \int \frac{x^2}{(x^2+2)^{\frac{3}{2}}} \, dx $, you can use the trigonometric substitution method. Let $ x = \sqrt{2} \tan(\theta) $, then $ dx = \sqrt{2} \sec^2(\theta) \, d\theta $. After substituting, you'll get an integral in terms of $ \theta $ that can be integrated easily.

How do you find the integral #intx^2/((x^2+2)^(3/2))dx# ?