Question

Evaluate the following integral
$\int{\dfrac{dx}{x({{x}^{3}}+1)}}$

Answer 1

The denominator of the integrand $\dfrac{1}{x({{x}^{3}}+1)}$ has 2 parts $x$ and $({{x}^{3}}+1)$ try to break the whole integrand into 2 parts , one having denominator $x$ and other $({{x}^{3}}+1)$ and then use the property of integral $\int{[f(x)+g(x)]dx=\int{f(x)dx+\int{g(x)dx}}}$ and evaluate the integral separately.

Complete step by step answer:
First we will consider the integrand and will try to break it into 2 parts , one having denominator $x$ and other having ${{x}^{3}}+1$ ,
Consider the given integrand,
$\dfrac{1}{x({{x}^{3}}+1)}$
Since, we are going to break it in 2 parts so consider the most general form of the 2 parts , i.e. $\dfrac{1}{x({{x}^{3}}+1)}=\dfrac{A}{x}+\dfrac{B{{x}^{2}}+C}{{{x}^{3}}+1}$
and find the values of $A,B$ and $C$
$\begin{aligned} & \dfrac{1}{x({{x}^{3}}+1)}=\dfrac{A{{x}^{3}}+A+B{{x}^{3}}+Cx}{x({{x}^{3}}+1)} \\\ & =\dfrac{(A+B){{x}^{3}}+A+Cx}{x({{x}^{3}}+1)} \end{aligned}$
Now, we will compare the numerators on both sides, and get equations to solve for $A,B$ and $C$ $\begin{aligned} & A+B=0 \\\ & C=0 \\\ & A=1 \end{aligned}$
By , solving these 3 equations , we can easily get,
$A=1,B=-1$ and $C=0$
Now , putting these values of $A,B$ and $C$ , the integrand becomes,
$\dfrac{1}{x({{x}^{3}}+1)}=\dfrac{1}{x}-\dfrac{{{x}^{2}}}{{{x}^{3}}+1}$
Now , consider the given integration and put the new integrand in it $\int{\dfrac{1}{x({{x}^{3}}+1)}dx=\int{\left( \dfrac{1}{x}-\dfrac{{{x}^{2}}}{{{x}^{3}}+1} \right)dx}}$
We know the property of integration,
$\int{[f(x)+g(x)]dx=\int{f(x)dx+\int{g(x)dx}}}$
Applying this property in the last step, we get,
$\begin{aligned} & =\int{\dfrac{1}{x}dx-\int{\dfrac{{{x}^{2}}}{{{x}^{3}}+1}dx}} \\\ & =\log (x)-\int{\dfrac{{{x}^{2}}}{{{x}^{3}}+1}dx} \end{aligned}$
Let, ${{x}^{3}}+1=t$
Differentiating both sides with respect to $''x''$ , we get
$\begin{aligned} & 3{{x}^{2}}dx=dt \\\ & {{x}^{2}}dx=\dfrac{1}{3}dt \end{aligned}$
Now, replacing $x$ in terms of $''t''$ in the integration on the last step , we get
$=\log (x)-\int{\dfrac{1}{t}\left( \dfrac{1}{3} \right)dt}$
Now, since constant can be taken out of the integration, we ge
$\begin{aligned} & =\log (x)-\dfrac{1}{3}\int{\dfrac{1}{t}dt} \\\ & =\log (x)-\dfrac{1}{3}\log (t) \\\ & \\\ \end{aligned}$
Now, putting the value of $''t''$ , we get

& =\log (x)-\dfrac{1}{3}\log ({{x}^{3}}+1) \\\ & =\dfrac{3}{3}\log (x)-\dfrac{1}{3}\log ({{x}^{3}}+1) \end{aligned}$$ We know from the property of log that, $\log (f{{(x)}^{a}})=a\log (f(x))$ Using this property of log in the step , we get $\begin{aligned} & =\dfrac{1}{3}\log ({{x}^{3}})-\dfrac{1}{3}\log ({{x}^{3}}+1) \\\ & \\\ \end{aligned}$ We know by the property of log that, $\log (f(x)-g(x))=\log \left( \dfrac{f(x)}{g(x)} \right)$ Using this property of log in the last step , we get, $=\dfrac{1}{3}\left( \log \dfrac{{{x}^{3}}}{{{x}^{3}}+1} \right)$ (or) $=\dfrac{1}{3}\log \left( \dfrac{1}{1+{{x}^{-3}}} \right)$ **Note:** The possibility of mistake here is that while dividing integrand into 2 parts , we have to put the most general form of the numerator , and we can get confused while choosing the most general form. The most general form of the numerator involves the degree 1 less than the denominator , if you take degree equal or greater than the degree of denominator, then it can further be reduced to the same format and if you take lesser degree then it may happen that you won’t get solution for the constants.