Final answer to the problem
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How should I solve this problem?
- Choose an option
- Integrate by partial fractions
- Integrate by substitution
- 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
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We can solve the integral $\int\mathrm{coth}\left(3x\right)dx$ 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 $3x$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part
Differentiate both sides of the equation $u=3x$
Find the derivative
The derivative of the linear function times a constant, is equal to the constant
The derivative of the linear function is equal to $1$
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
Rearrange the equation
Divide both sides of the equation by $3$
Isolate $dx$ in the previous equation
Substituting $u$ and $dx$ in the integral and simplify
Take the constant $\frac{1}{3}$ out of the integral
We can solve the integral $\int\mathrm{coth}\left(u\right)du$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula
Taking the derivative of hyperbolic cotangent
The derivative of the linear function is equal to $1$
First, identify or choose $u$ and calculate it's derivative, $du$
Now, identify $dv$ and calculate $v$
Solve the integral to find $v$
The integral of a constant is equal to the constant times the integral's variable
The integral of a function times a constant ($-1$) is equal to the constant times the integral of the function
Any expression multiplied by $1$ is equal to itself
Now replace the values of $u$, $du$ and $v$ in the last formula
Multiply the single term $\frac{1}{3}$ by each term of the polynomial $\left(u\mathrm{coth}\left(u\right)+\int u\mathrm{csch}\left(u\right)^2du\right)$
Replace $u$ with the value that we assigned to it in the beginning: $3x$
Replace $u$ with the value that we assigned to it in the beginning: $3x$
Multiply the fraction and term in $3\cdot \frac{1}{3}x\mathrm{coth}\left(u\right)$
Replace $u$ with the value that we assigned to it in the beginning: $3x$
Replace $u$ with the value that we assigned to it in the beginning: $3x$
We can solve the integral $\int u\mathrm{csch}\left(u\right)^2du$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula
First, identify or choose $u$ and calculate it's derivative, $du$
Now, identify $dv$ and calculate $v$
Solve the integral to find $v$
Apply the formula: $\int\mathrm{csch}\left(\theta \right)^2dx$$=-\mathrm{coth}\left(\theta \right)+C$, where $x=u$
Now replace the values of $u$, $du$ and $v$ in the last formula
Multiply the single term $\frac{1}{3}$ by each term of the polynomial $\left(-u\mathrm{coth}\left(u\right)+\int\mathrm{coth}\left(u\right)du\right)$
Replace $u$ with the value that we assigned to it in the beginning: $3x$
Multiply the fraction and term in $3-\frac{1}{3}x\mathrm{coth}\left(3x\right)$
Apply the trigonometric identity: $\mathrm{coth}\left(\theta \right)$$=\frac{\mathrm{cosh}\left(\theta \right)}{\mathrm{sinh}\left(\theta \right)}$, where $x=u$
We can solve the integral $\int\frac{\mathrm{cosh}\left(u\right)}{\mathrm{sinh}\left(u\right)}du$ 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 $\mathrm{sinh}\left(u\right)$ it's a good candidate for substitution. Let's define a variable $v$ and assign it to the choosen part
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
Isolate $du$ in the previous equation
Substituting $v$ and $du$ in the integral and simplify
The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$
Replace $v$ with the value that we assigned to it in the beginning: $\mathrm{sinh}\left(u\right)$
Replace $u$ with the value that we assigned to it in the beginning: $3x$
The integral $\frac{1}{3}\int u\mathrm{csch}\left(u\right)^2du$ results in: $-x\mathrm{coth}\left(3x\right)+\frac{1}{3}\ln\left(\mathrm{sinh}\left(3x\right)\right)$
Gather the results of all integrals
Cancel like terms $x\mathrm{coth}\left(3x\right)$ and $-x\mathrm{coth}\left(3x\right)$
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$