Question

Calculate the following integral: $\int{\dfrac{dx}{x({{x}^{5}}+3)}}$

Answer 1

We will start with multiplying x inside the bracket in the denominator and then we will take ${{x}^{6}}$outside, after that, we can substitute $u=1+\dfrac{3}{{{x}^{5}}}$ and find $\dfrac{du}{dx}$, by this, we will expand the function into a simpler form and then we will apply the u-substitution method or the reverse chain rule to do the integration. Further, we will apply some logarithmic properties in order to simplify the solution.

Complete step-by-step solution:
We have with us the following integral: $\int{\dfrac{dx}{x({{x}^{5}}+3)}}$, We will now multiply the x in the denominator into the bracket.
And after multiplying, let’s rewrite the given integral as: $\int{\dfrac{dx}{({{x}^{6}}+3x)}}$
After this, we will expand the fraction by $\dfrac{1}{{{x}^{6}}}$, this we will do by multiplying and dividing the denominator by ${{x}^{6}}$.

& \Rightarrow \int{\dfrac{dx}{\dfrac{{{x}^{6}}}{{{x}^{6}}}\left( {{x}^{6}}+3x \right)}}=\int{\dfrac{dx}{{{x}^{6}}\left( \dfrac{{{x}^{6}}}{{{x}^{6}}}+\dfrac{3x}{{{x}^{6}}} \right)}} \\\ & \Rightarrow \int{\dfrac{dx}{{{x}^{6}}\left( 1+\dfrac{3}{{{x}^{5}}} \right)}}\text{ }.................................\text{Equation 1}\text{.} \\\ & \\\ \end{aligned}$$ Now, we will apply the u-substitution method or the reverse chain rule, in this rule we substitute the complex function with a single variable and then differentiate to obtain the derivate to avoid complexity and then again re-substitute the whole value. So here, we will replace $1+\dfrac{3}{{{x}^{5}}}\to u$ Therefore, when $u=1+\dfrac{3}{{{x}^{5}}}$, differentiating both sides to obtain the value of du. We will use the power rule here for differentiation, as we know the power rule is : $\left( \dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}} \right)$ So, we can apply it as shown below: $$\begin{aligned} & \Rightarrow u=(1+3{{x}^{-5}}) \\\ & \Rightarrow du=\left( 0+\left( 3\times -5{{x}^{(-5-1)}} \right) \right)dx \\\ & \Rightarrow du=-15{{x}^{-6}}dx \\\ & \Rightarrow du=\dfrac{-15}{{{x}^{6}}}dx \\\ & \Rightarrow dx=\dfrac{{{x}^{6}}}{-15}du \\\ \end{aligned}$$ Now substituting the obtained values of $u=1+\dfrac{3}{{{x}^{5}}},dx=-\dfrac{{{x}^{6}}}{15}du$ in equation-1 We have equation 1 as $$\int{\dfrac{dx}{{{x}^{6}}\left( 1+\dfrac{3}{{{x}^{5}}} \right)}}$$ ; After replacement we get the following equation: $$\int{\dfrac{-{{x}^{6}}du}{15{{x}^{6}}.u}}$$ (Cancelling ${{x}^{6}}$ ) $\Rightarrow \int{\dfrac{-du}{15u}}$ We will take the constant out : $\int{a}\cdot f\left( x \right)dx=a\cdot \int{f}\left( x \right)dx$ $\int{\dfrac{-du}{15u}}=-\dfrac{1}{15}\int{\dfrac{du}{u}}.......................\text{ Equation 2}\text{.}$ Now we will solve: $$\int{\dfrac{1}{u}du}$$ (This is a standard integral : $\int{\dfrac{1}{x}dx=\ln \left( x \right)}+C$ ) Therefore, we get the following: $\int{\dfrac{1}{u}du=\ln \left( u \right)}+C$ Where C is a constant. So $-\dfrac{1}{15}\int{\dfrac{du}{u}}=-\dfrac{1}{15}\ln \left( u \right)+C$ After re-substituting the value of u into the above equation, we will have: $\dfrac{-1}{15}\ln \left( 1+\dfrac{3}{{{x}^{5}}} \right)+C$ On further simplifying we will get: $\begin{aligned} & \dfrac{-1}{15}\ln \left( 1+\dfrac{3}{{{x}^{5}}} \right)+C=\dfrac{\ln \left( 1+\dfrac{3}{{{x}^{5}}} \right)}{-15}+C\Rightarrow \dfrac{\ln \left( \dfrac{{{x}^{5}}+3}{{{x}^{5}}} \right)}{-15}+C \\\ & \\\ \end{aligned}$ Applying the logarithm property further: $\log \left( \dfrac{f\left( x \right)}{g\left( x \right)} \right)=\log f\left( x \right)-\log g\left( x \right)$ $\dfrac{\ln \left( \dfrac{{{x}^{5}}+3}{{{x}^{5}}} \right)}{-15}+C=\dfrac{\ln \left( {{x}^{5}}+3 \right)-\ln \left( {{x}^{5}} \right)}{-15}+C$ { Now we will apply: $\log f{{\left( x \right)}^{n}}=n\log f\left( x \right)$ } $$\begin{aligned} & -\dfrac{\ln \left( {{x}^{5}}+3 \right)-\ln \left( {{x}^{5}} \right)}{15}+C=\dfrac{5\ln \left( x \right)-\ln \left( {{x}^{5}}+3 \right)}{15}+C \\\ & \Rightarrow \dfrac{\ln \left( x \right)}{3}-\dfrac{\ln \left( {{x}^{5}}+3 \right)}{15}+C \\\ \end{aligned}$$ We will apply the absolute value function to logarithm functions in order to extend the integral’s domain: **Hence, $$\int{\dfrac{dx}{x({{x}^{5}}+3)}}=\dfrac{\ln \left( \left| x \right| \right)}{3}-\dfrac{\ln \left( \left| {{x}^{5}}+3 \right| \right)}{15}+C$$** **Note:** This integration can also be solved by the partial-fraction decomposition method. In Partial-fraction decomposition first, we start with the simplified answer and then taking it back apart, then we decompose the final expression into the initial polynomial fractions. For decomposition first, we write the function in the form of fractions where the denominators are factors of the given function and since the numerator is unknown we randomly assign them with any variable. Remember to use the absolute value of log functions because we want to cover the whole domain.