Find the integral $\int\frac{16}{x^2\sqrt{x^2+9}}dx$

Step-by-step Solution

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Final answer to the problem

$\frac{16\sqrt{x^2+9}}{-9x}+C_0$
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Step-by-step Solution

How should I solve this problem?

  • Integrate by substitution
  • Integrate by partial fractions
  • Integrate by parts
  • Integrate using tabular integration
  • Integrate by trigonometric substitution
  • Weierstrass Substitution
  • Integrate using trigonometric identities
  • Integrate using basic integrals
  • Product of Binomials with Common Term
  • FOIL Method
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1

We can solve the integral $\int\frac{16}{x^2\sqrt{x^2+9}}dx$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $x^2$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=x^2$
2

Now, in order to rewrite $dx$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above

$du=2xdx$
3

Isolate $dx$ in the previous equation

$\frac{du}{2x}=dx$
4

Rewriting $x$ in terms of $u$

$x=\sqrt{u}$
5

Substituting $u$, $dx$ and $x$ in the integral and simplify

$\int\frac{8}{\sqrt{u^{3}}\sqrt{u+9}}du$
6

We can solve the integral $\int\frac{8}{\sqrt{u^{3}}\sqrt{u+9}}du$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $v$), which when substituted makes the integral easier. We see that $\sqrt{u+9}$ it's a good candidate for substitution. Let's define a variable $v$ and assign it to the choosen part

$v=\sqrt{u+9}$
7

Now, in order to rewrite $du$ in terms of $dv$, we need to find the derivative of $v$. We need to calculate $dv$, we can do that by deriving the equation above

$dv=\frac{1}{2}\left(u+9\right)^{-\frac{1}{2}}du$
8

Isolate $du$ in the previous equation

$\frac{dv}{\frac{1}{2}\left(u+9\right)^{-\frac{1}{2}}}=du$
9

Rewriting $u$ in terms of $v$

$u=v^2-9$
10

Substituting $v$, $du$ and $u$ in the integral and simplify

$\int\frac{16}{\sqrt{\left(v^2-9\right)^{3}}}dv$
11

We can solve the integral $\int\frac{16}{\sqrt{\left(v^2-9\right)^{3}}}dv$ by applying integration method of trigonometric substitution using the substitution

$v=3\sec\left(\theta \right)$
12

Now, in order to rewrite $d\theta$ in terms of $dv$, we need to find the derivative of $v$. We need to calculate $dv$, we can do that by deriving the equation above

$dv=3\sec\left(\theta \right)\tan\left(\theta \right)d\theta$
13

Substituting in the original integral, we get

$\int\frac{48\sec\left(\theta \right)\tan\left(\theta \right)}{\sqrt{\left(9\sec\left(\theta \right)^2-9\right)^{3}}}d\theta$
14

Factor the polynomial $9\sec\left(\theta \right)^2-9$ by it's greatest common factor (GCF): $9$

$\int\frac{48\sec\left(\theta \right)\tan\left(\theta \right)}{\sqrt{\left(9\left(\sec\left(\theta \right)^2-1\right)\right)^{3}}}d\theta$
15

The power of a product is equal to the product of it's factors raised to the same power

$\int\frac{48\sec\left(\theta \right)\tan\left(\theta \right)}{27\sqrt{\left(\sec\left(\theta \right)^2-1\right)^{3}}}d\theta$
16

Apply the trigonometric identity: $\sec\left(\theta \right)^2-1$$=\tan\left(\theta \right)^2$, where $x=\theta $

$\int\frac{48\sec\left(\theta \right)\tan\left(\theta \right)}{27\sqrt{\left(\tan\left(\theta \right)^2\right)^{3}}}d\theta$
17

Taking the constant ($48$) out of the integral

$48\int\frac{\sec\left(\theta \right)\tan\left(\theta \right)}{27\sqrt{\left(\tan\left(\theta \right)^2\right)^{3}}}d\theta$
18

Simplify $\sqrt{\left(\tan\left(\theta \right)^2\right)^{3}}$ using the power of a power property: $\left(a^m\right)^n=a^{m\cdot n}$. In the expression, $m$ equals $2$ and $n$ equals $\frac{3}{2}$

$48\int\frac{\sec\left(\theta \right)\tan\left(\theta \right)}{27\tan\left(\theta \right)^{3}}d\theta$
19

Simplify the fraction by $\tan\left(\theta \right)$

$48\int\frac{\sec\left(\theta \right)}{27\tan\left(\theta \right)^{2}}d\theta$
20

Take the constant $\frac{1}{27}$ out of the integral

$48\cdot \left(\frac{1}{27}\right)\int\frac{\sec\left(\theta \right)}{\tan\left(\theta \right)^{2}}d\theta$
21

Simplify the expression

$\frac{16}{9}\int\frac{\cos\left(\theta \right)}{\sin\left(\theta \right)^{2}}d\theta$
22

We can solve the integral $\int\frac{\cos\left(\theta \right)}{\sin\left(\theta \right)^{2}}d\theta$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $w$), which when substituted makes the integral easier. We see that $\sin\left(\theta \right)$ it's a good candidate for substitution. Let's define a variable $w$ and assign it to the choosen part

$w=\sin\left(\theta \right)$
23

Now, in order to rewrite $d\theta$ in terms of $dw$, we need to find the derivative of $w$. We need to calculate $dw$, we can do that by deriving the equation above

$dw=\cos\left(\theta \right)d\theta$
24

Isolate $d\theta$ in the previous equation

$\frac{dw}{\cos\left(\theta \right)}=d\theta$
25

Substituting $w$ and $d\theta$ in the integral and simplify

$\frac{16}{9}\int\frac{1}{w^{2}}dw$
26

Rewrite the exponent using the power rule $\frac{a^m}{a^n}=a^{m-n}$, where in this case $m=0$

$\frac{16}{9}\int w^{-2}dw$
27

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, such as $-2$

$\frac{16}{9}\frac{w^{-1}}{-1}$
28

Simplify the expression

$\frac{16}{-9w}$
29

Replace $w$ with the value that we assigned to it in the beginning: $\sin\left(\theta \right)$

$\frac{16}{-9\sin\left(\theta \right)}$
30

Express the variable $\theta$ in terms of the original variable $x$

$\frac{16v}{-9\sqrt{v^2-9}}$
31

Replace $v$ with the value that we assigned to it in the beginning: $\sqrt{u+9}$

$\frac{16\sqrt{u+9}}{-9\sqrt{\left(\sqrt{u+9}\right)^2-9}}$
32

Replace $u$ with the value that we assigned to it in the beginning: $x^2$

$\frac{16\sqrt{x^2+9}}{-9\sqrt{x^2}}$
33

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$\frac{16\sqrt{x^2+9}}{-9x}+C_0$

Final answer to the problem

$\frac{16\sqrt{x^2+9}}{-9x}+C_0$

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Plotting: $\frac{16\sqrt{x^2+9}}{-9x}+C_0$

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a
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g
m
n
u
v
w
x
y
z
.
(◻)
+
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×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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