Final Answer
$\frac{\mathrm{arcsec}\left(\frac{x}{5}\right)}{10}+\frac{-\sqrt{x^2-25}}{2x^2}+C_0$
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Step-by-step Solution
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1
We can solve the integral $\int\frac{\sqrt{x^2-25}}{x^3}dx$ by applying integration method of trigonometric substitution using the substitution
$x=5\sec\left(\theta \right)$
Intermediate steps
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
$dx=5\sec\left(\theta \right)\tan\left(\theta \right)d\theta$
Explain this step further
3
Substituting in the original integral, we get
$\int\frac{\frac{1}{25}\sqrt{25\sec\left(\theta \right)^2-25}\sec\left(\theta \right)\tan\left(\theta \right)}{\sec\left(\theta \right)^3}d\theta$
Intermediate steps
$\int\frac{\sqrt{25\sec\left(\theta \right)^2-25}\tan\left(\theta \right)}{25\sec\left(\theta \right)^{2}}d\theta$
Explain this step further
5
Factor the polynomial $25\sec\left(\theta \right)^2-25$ by it's greatest common factor (GCF): $25$
$\int\frac{\sqrt{25\left(\sec\left(\theta \right)^2-1\right)}\tan\left(\theta \right)}{25\sec\left(\theta \right)^{2}}d\theta$
6
The power of a product is equal to the product of it's factors raised to the same power
$\int\frac{5\sqrt{\sec\left(\theta \right)^2-1}\tan\left(\theta \right)}{25\sec\left(\theta \right)^{2}}d\theta$
7
Apply the trigonometric identity: $\sec\left(\theta \right)^2-1$$=\tan\left(\theta \right)^2$, where $x=\theta $
$\int\frac{5\sqrt{\tan\left(\theta \right)^2}\tan\left(\theta \right)}{25\sec\left(\theta \right)^{2}}d\theta$
8
Taking the constant ($5$) out of the integral
$5\int\frac{\sqrt{\tan\left(\theta \right)^2}\tan\left(\theta \right)}{25\sec\left(\theta \right)^{2}}d\theta$
9
Simplify $\sqrt{\tan\left(\theta \right)^2}$ using the power of a power property: $\left(a^m\right)^n=a^{m\cdot n}$. In the expression, $m$ equals $2$ and $n$ equals $\frac{1}{2}$
$5\int\frac{\tan\left(\theta \right)\tan\left(\theta \right)}{25\sec\left(\theta \right)^{2}}d\theta$
10
When multiplying two powers that have the same base ($\tan\left(\theta \right)$), you can add the exponents
$5\int\frac{\tan\left(\theta \right)^2}{25\sec\left(\theta \right)^{2}}d\theta$
11
Apply the trigonometric identity: $\frac{\tan\left(\theta \right)^n}{\sec\left(\theta \right)^n}$$=\sin\left(\theta \right)^n$, where $x=\theta $ and $n=2$
$5\int\frac{\sin\left(\theta \right)^2}{25}d\theta$
12
Take the constant $\frac{1}{25}$ out of the integral
$5\cdot \left(\frac{1}{25}\right)\int\sin\left(\theta \right)^2d\theta$
Intermediate steps
13
Simplify the expression inside the integral
$\frac{1}{5}\int\sin\left(\theta \right)^2d\theta$
Explain this step further
14
Apply the formula: $\int\sin\left(\theta \right)^2dx$$=\frac{\theta }{2}-\frac{1}{4}\sin\left(2\theta \right)+C$, where $x=\theta $
$\frac{1}{5}\left(\frac{\theta }{2}-\frac{1}{4}\sin\left(2\theta \right)\right)$
15
Express the variable $\theta$ in terms of the original variable $x$
$\frac{1}{5}\left(\frac{\mathrm{arcsec}\left(\frac{x}{5}\right)}{2}-\frac{1}{4}\sin\left(2\theta \right)\right)$
16
Using the sine double-angle identity: $\sin\left(2\theta\right)=2\sin\left(\theta\right)\cos\left(\theta\right)$
$\frac{1}{5}\left(\frac{\mathrm{arcsec}\left(\frac{x}{5}\right)}{2}-\frac{1}{2}\sin\left(\theta \right)\cos\left(\theta \right)\right)$
Why does sin(2x) = 2sin(x)cos(x) ?
Intermediate steps
17
Express the variable $\theta$ in terms of the original variable $x$
$\frac{1}{5}\left(\frac{\mathrm{arcsec}\left(\frac{x}{5}\right)}{2}+\frac{-5\sqrt{x^2-25}}{2x^2}\right)$
Explain this step further
18
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
$\frac{1}{5}\left(\frac{\mathrm{arcsec}\left(\frac{x}{5}\right)}{2}+\frac{-5\sqrt{x^2-25}}{2x^2}\right)+C_0$
Intermediate steps
$\frac{\mathrm{arcsec}\left(\frac{x}{5}\right)}{10}+\frac{-\sqrt{x^2-25}}{2x^2}+C_0$
Explain this step further
Final Answer$\frac{\mathrm{arcsec}\left(\frac{x}{5}\right)}{10}+\frac{-\sqrt{x^2-25}}{2x^2}+C_0$