How do you find the integral
#intx^3/(sqrt(x^2+9))dx# ?

Question

How do you find the integral
 #intx^3/(sqrt(x^2+9))dx# ?

Emilia Aguilar · Accepted Answer

#=(x^2+9)^(3/2)/3-9sqrt(x^2+9)+c#, where #c# is a constant

Explanation :

#=intx^3/sqrt(x^2+9)dx#

Using Integration by Substitution,

Let's assume #x^2+9=t^2#, then

#2xdx=2tdt#, #=>xdx=tdt#

#=int((t^2-9)t)/tdt#

#=intt^2dt-int9dt#

#=t^3/3-9t+c#, where #c# is a constant

Substituting #t# back,

#=(x^2+9)^(3/2)/3-9sqrt(x^2+9)+c#, where #c# is a constant

Luke Alvarez · Answer

undefined To find the integral $ \int \frac{x^3}{\sqrt{x^2 + 9}} \, dx $, we can use the trigonometric substitution method.

We let $ x = 3 \tan(\theta) $, then $ dx = 3 \sec^2(\theta) \, d\theta $.

Substituting $ x $ and $ dx $ into the integral, we get:

$$ \int \frac{(3 \tan(\theta))^3}{\sqrt{(3 \tan(\theta))^2 + 9}} \cdot 3 \sec^2(\theta) \, d\theta $$

Simplifying, we have:

$$ \int \frac{27 \tan^3(\theta)}{\sqrt{9 \tan^2(\theta) + 9}} \cdot 3 \sec^2(\theta) \, d\theta $$

$$ = 27 \int \frac{\tan^3(\theta)}{\sqrt{9(\tan^2(\theta) + 1)}} \cdot \sec^2(\theta) \, d\theta $$

$$ = 27 \int \frac{\tan^3(\theta)}{\sqrt{9 \sec^2(\theta)}} \cdot \sec^2(\theta) \, d\theta $$

$$ = 27 \int \frac{\tan^3(\theta)}{3 \sec(\theta)} \cdot \sec^2(\theta) \, d\theta $$

$$ = 9 \int \tan^3(\theta) \, d\theta $$

$$ = 9 \int (\sec^2(\theta) - \sec(\theta)) \tan(\theta) \, d\theta $$

$$ = 9 \left(\frac{\sec^3(\theta)}{3} - \ln|\sec(\theta) + \tan(\theta)|\right) + C $$

$$ = 3\sec^3(\theta) - 9 \ln|\sec(\theta) + \tan(\theta)| + C $$

Finally, we substitute back $ x = 3 \tan(\theta) $ to get the answer in terms of $ x $:

$$ = 3x^3 - 9 \ln|x + \sqrt{x^2 + 9}| + C $$