Exercise$\int\left(\frac{1}{x^5\sqrt{9x^2-1}}\right)dx$
Step-by-step Solution
1
First, factor the terms inside the radical by $9$ for an easier handling
$\int\frac{1}{x^5\sqrt{9\left(x^2-\frac{1}{9}\right)}}dx$
2
Taking the constant out of the radical
$\int\frac{1}{3x^5\sqrt{x^2-\frac{1}{9}}}dx$
3
We can solve the integral $\int\frac{1}{3x^5\sqrt{x^2-\frac{1}{9}}}dx$ by applying integration method of trigonometric substitution using the substitution
$x=\frac{1}{3}\sec\left(\theta \right)$
Intermediate steps
4
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
$dx=\frac{1}{3}\sec\left(\theta \right)\tan\left(\theta \right)d\theta$
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Intermediate steps
5
Substituting in the original integral, we get
$\int\frac{1}{3}\frac{1}{\frac{1}{81}\sec\left(\theta \right)^5\sqrt{\frac{1}{9}\sec\left(\theta \right)^2-\frac{1}{9}}}\sec\left(\theta \right)\tan\left(\theta \right)d\theta$
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Intermediate steps
$\int\frac{27\tan\left(\theta \right)}{\sec\left(\theta \right)^{4}\sqrt{\frac{1}{9}\sec\left(\theta \right)^2-\frac{1}{9}}}d\theta$
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7
Taking the constant ($27$) out of the integral
$27\int\frac{\tan\left(\theta \right)}{\sec\left(\theta \right)^{4}\sqrt{\frac{1}{9}\sec\left(\theta \right)^2-\frac{1}{9}}}d\theta$
Intermediate steps
8
Rewrite the trigonometric expression $\frac{\tan\left(\theta \right)}{\sec\left(\theta \right)^{4}\sqrt{\frac{1}{9}\sec\left(\theta \right)^2-\frac{1}{9}}}$ inside the integral
$27\int\frac{3\tan\left(\theta \right)\cos\left(\theta \right)^{5}}{\sin\left(\theta \right)}d\theta$
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Intermediate steps
9
Simplify the expression
$81\int\cos\left(\theta \right)^{4}d\theta$
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10
Apply the formula: $\int\cos\left(\theta \right)^ndx$$=\frac{\cos\left(\theta \right)^{\left(n-1\right)}\sin\left(\theta \right)}{n}+\frac{n-1}{n}\int\cos\left(\theta \right)^{\left(n-2\right)}dx$, where $x=\theta $ and $n=4$
$81\left(\frac{\cos\left(\theta \right)^{3}\sin\left(\theta \right)}{4}+\frac{3}{4}\int\cos\left(\theta \right)^{2}d\theta\right)$
11
Solve the product $81\left(\frac{\cos\left(\theta \right)^{3}\sin\left(\theta \right)}{4}+\frac{3}{4}\int\cos\left(\theta \right)^{2}d\theta\right)$
$\frac{81\cos\left(\theta \right)^{3}\sin\left(\theta \right)}{4}+\frac{243}{4}\int\cos\left(\theta \right)^{2}d\theta$
Intermediate steps
12
Express the variable $\theta$ in terms of the original variable $x$
$\frac{\frac{3\sqrt{x^2-\frac{1}{9}}}{x^{3}x}}{4}+\frac{243}{4}\int\cos\left(\theta \right)^{2}d\theta$
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Intermediate steps
13
Simplify the expression
$\frac{3\sqrt{x^2-\frac{1}{9}}}{4x^{4}}+\frac{243}{4}\int\cos\left(\theta \right)^{2}d\theta$
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Intermediate steps
14
The integral $\frac{243}{4}\int\cos\left(\theta \right)^{2}d\theta$ results in: $\frac{243}{4}\left(\frac{1}{2}\mathrm{arcsec}\left(3x\right)+\frac{\frac{1}{3}\sqrt{x^2-\frac{1}{9}}}{2x^2}\right)$
$\frac{243}{4}\left(\frac{1}{2}\mathrm{arcsec}\left(3x\right)+\frac{\frac{1}{3}\sqrt{x^2-\frac{1}{9}}}{2x^2}\right)$
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15
Gather the results of all integrals
$\frac{3\sqrt{x^2-\frac{1}{9}}}{4x^{4}}+\frac{243}{4}\left(\frac{1}{2}\mathrm{arcsec}\left(3x\right)+\frac{\frac{1}{3}\sqrt{x^2-\frac{1}{9}}}{2x^2}\right)$
16
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
$\frac{3\sqrt{x^2-\frac{1}{9}}}{4x^{4}}+\frac{243}{4}\left(\frac{1}{2}\mathrm{arcsec}\left(3x\right)+\frac{\frac{1}{3}\sqrt{x^2-\frac{1}{9}}}{2x^2}\right)+C_0$
Intermediate steps
$\frac{3\sqrt{x^2-\frac{1}{9}}}{4x^{4}}+\frac{243}{8}\mathrm{arcsec}\left(3x\right)+\frac{81\sqrt{x^2-\frac{1}{9}}}{8x^2}+C_0$
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Final answer to the exercise $\frac{3\sqrt{x^2-\frac{1}{9}}}{4x^{4}}+\frac{243}{8}\mathrm{arcsec}\left(3x\right)+\frac{81\sqrt{x^2-\frac{1}{9}}}{8x^2}+C_0$