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

Step-by-step Solution

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Final Answer

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

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1

We can solve the integral $\int\frac{16}{x^2\sqrt{x^2+9}}dx$ by applying integration method of trigonometric substitution using the substitution

$x=3\tan\left(\theta \right)$

Differentiate both sides of the equation $x=3\tan\left(\theta \right)$

$dx=\frac{d}{d\theta}\left(3\tan\left(\theta \right)\right)$

Find the derivative

$\frac{d}{d\theta}\left(3\tan\left(\theta \right)\right)$

The derivative of a function multiplied by a constant ($3$) is equal to the constant times the derivative of the function

$3\frac{d}{d\theta}\left(\tan\left(\theta \right)\right)$

The derivative of the tangent of a function is equal to secant squared of that function times the derivative of that function, in other words, if ${f(x) = tan(x)}$, then ${f'(x) = sec^2(x)\cdot D_x(x)}$

$3\frac{d}{d\theta}\left(\theta \right)\sec\left(\theta \right)^2$

The derivative of the linear function is equal to $1$

$3\sec\left(\theta \right)^2$
2

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

$dx=3\sec\left(\theta \right)^2d\theta$
3

Substituting in the original integral, we get

$\int\frac{\frac{16}{3}\sec\left(\theta \right)^2}{\tan\left(\theta \right)^2\sqrt{9\tan\left(\theta \right)^2+9}}d\theta$

Rewrite the fraction $\frac{\frac{16}{3}\sec\left(\theta \right)^2}{\tan\left(\theta \right)^2\sqrt{9\tan\left(\theta \right)^2+9}}$

$\int\frac{16}{3\tan\left(\theta \right)^2\sqrt{9\tan\left(\theta \right)^2+9}}\sec\left(\theta \right)^2d\theta$

Multiplying the fraction by $\sec\left(\theta \right)^2$

$\int\frac{16\sec\left(\theta \right)^2}{3\tan\left(\theta \right)^2\sqrt{9\tan\left(\theta \right)^2+9}}d\theta$
4

Simplifying

$\int\frac{16\sec\left(\theta \right)^2}{3\tan\left(\theta \right)^2\sqrt{9\tan\left(\theta \right)^2+9}}d\theta$
5

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

$\int\frac{16\sec\left(\theta \right)^2}{3\tan\left(\theta \right)^2\sqrt{9\left(\tan\left(\theta \right)^2+1\right)}}d\theta$
6

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

$\int\frac{16\sec\left(\theta \right)^2}{9\tan\left(\theta \right)^2\sqrt{\tan\left(\theta \right)^2+1}}d\theta$
7

Applying the trigonometric identity: $1+\tan\left(\theta \right)^2 = \sec\left(\theta \right)^2$

$\int\frac{16\sec\left(\theta \right)^2}{9\tan\left(\theta \right)^2\sqrt{\sec\left(\theta \right)^2}}d\theta$
Why is tan(x)^2+1 = sec(x)^2 ?
8

Taking the constant ($16$) out of the integral

$16\int\frac{\sec\left(\theta \right)^2}{9\tan\left(\theta \right)^2\sqrt{\sec\left(\theta \right)^2}}d\theta$
9

Simplify $\sqrt{\sec\left(\theta \right)^2}$ 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 $\frac{1}{2}$

$16\int\frac{\sec\left(\theta \right)^2}{9\tan\left(\theta \right)^2\sec\left(\theta \right)}d\theta$
10

Simplify the fraction $\frac{\sec\left(\theta \right)^2}{9\tan\left(\theta \right)^2\sec\left(\theta \right)}$ by $\sec\left(\theta \right)$

$16\int\frac{\sec\left(\theta \right)}{9\tan\left(\theta \right)^2}d\theta$
11

Take the constant $\frac{1}{9}$ out of the integral

$16\cdot \frac{1}{9}\int\frac{\sec\left(\theta \right)}{\tan\left(\theta \right)^2}d\theta$

Multiply $16$ times $\frac{1}{9}$

$\frac{16}{9}\int\frac{\sec\left(\theta \right)}{\tan\left(\theta \right)^2}d\theta$

Apply the trigonometric identity: $\frac{\sec\left(\theta \right)}{\tan\left(\theta \right)^m}$$=\frac{\cos\left(\theta \right)^{\left(m-1\right)}}{\sin\left(\theta \right)^m}$, where $x=\theta $ and $m=2$

$\frac{16}{9}\int\frac{\cos\left(\theta \right)}{\sin\left(\theta \right)^2}d\theta$
12

Simplify the expression inside the integral

$\frac{16}{9}\int\frac{\cos\left(\theta \right)}{\sin\left(\theta \right)^2}d\theta$
13

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 $u$), 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 $u$ and assign it to the choosen part

$u=\sin\left(\theta \right)$

Differentiate both sides of the equation $u=\sin\left(\theta \right)$

$du=\frac{d}{d\theta}\left(\sin\left(\theta \right)\right)$

Find the derivative

$\frac{d}{d\theta}\left(\sin\left(\theta \right)\right)$

The derivative of the sine of a function is equal to the cosine of that function times the derivative of that function, in other words, if ${f(x) = \sin(x)}$, then ${f'(x) = \cos(x)\cdot D_x(x)}$

$\cos\left(\theta \right)$
14

Now, in order to rewrite $d\theta$ 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

$du=\cos\left(\theta \right)d\theta$
15

Isolate $d\theta$ in the previous equation

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

Simplify the fraction $\frac{\frac{\cos\left(\theta \right)}{u^2}}{\cos\left(\theta \right)}$ by $\cos\left(\theta \right)$

$\int\frac{1}{u^2}du$
16

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

$\frac{16}{9}\int\frac{1}{u^2}du$
17

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

$\frac{16}{9}\int u^{-2}du$
18

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 $-2$

$-\frac{16}{9}u^{-1}$

$-\frac{16}{9}\sin\left(\theta \right)^{-1}$
19

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

$-\frac{16}{9}\sin\left(\theta \right)^{-1}$
20

Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number

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

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

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

Multiplying the fraction by $9$

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

Divide fractions $\frac{-16}{\frac{9x}{\sqrt{x^2+9}}}$ with Keep, Change, Flip: $a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}$

$\frac{-16\sqrt{x^2+9}}{9x}$
21

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

$\frac{-16\sqrt{x^2+9}}{9x}$
22

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

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

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Function Plot

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

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x
y
z
.
(◻)
+
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◻/◻
/
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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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