Question

please answer this question​

Answer 1

Answer:

Given Integral is:

$\\ { \displaystyle{ \int{ \large{ \rm{ \frac{1}{x( {x}^{3} + 1) }dx }}}}}$

Can be written as:

$\\ { \large{ \longrightarrow{ \displaystyle{ \int { \rm{\frac{3 {x}^{2} }{3 {x}^{2} \times x( {x}^{3} + 1) } dx}}}}}}$

$\\ { \large{ \longrightarrow \frac{1}{3} \: { \displaystyle{ \int{ \rm { \frac{ {3x}^{2} }{ {x}^{3} ( {x}^{3} + 1)} dx}}}}}}$

Now to evaluate this integral , we use the method of Substitution.

So, Substitute:

$\: \: \: \: \: \: \: \: \: \: \: \: { \rightarrow \: {\large{\rm{ {x}^{3} = y}}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \rightarrow \: { \large{ \rm{3 {x}^{2} dx = dy}}}$

So, on Substituting the values in above integral, we get:

$\\ { \large{ \longrightarrow{ \frac{1}{3} \: \displaystyle{ \int{\rm{ \frac{1}{y(y + 1)} }}}}}}$

$\\ { \large{ \longrightarrow{ \frac{1}{3} \: { \displaystyle{ \int{ \rm{ \frac{1 + y - y}{y(y + 1)}dy }}}}}}}$

$\\ { \large{ \longrightarrow{ \frac{1}{3} \: { \displaystyle{ \int{ {\rm{ \frac{y + 1 - y}{y(y + 1)}dy }}}}}}}}$

$\\ { \large{ \longrightarrow{ \frac{1}{3} \: { \displaystyle{ \int{ \rm{ \left[ \frac{1}{y} - \frac{1}{y + 1} \right ] }dy}}}}}}$

${ \large{ \longrightarrow{ \rm{ \frac{1}{3} \left( log |y| - log |y + 1| \right) + c}}}}$

${ \large{ \longrightarrow{ \rm{ \frac{1}{3} log | \frac{y}{y + 1} | + c }}}}$

${ \large{ \longrightarrow{ \rm{ \frac{1}{3} log | \frac{ {x}^{3} }{ {x}^{3} + 1 } | + c }}}}$

Hence,

$\\{ \boxed{ \pink { \large{ \displaystyle{ \int{ \rm{ \frac{dx}{x ({x}^{3 } + 1 )} = \frac{1}{3 } log{ \huge{ |}} \frac{ {x}^{3} }{ {x}^{3} + 1} { \huge{ | }} + c}}}}}}}$

Answer 2

After applying algebraic substitution, partial fraction decomposition and integration rules the integral of the non-trascendental expression $\frac{1}{x\cdot (x^{3}+1)}$ is equal to the trascendental expression $I = \frac{1}{3}\cdot \ln \left|\frac{x^{3}}{x^{3}+1} \right| + C$.

How to determine the indefinite integral of a given expression

According to the statement we have an integrable rational expression formed by non-trascendental expressions, that is, expressions whose behavior is described solely by algebraic components. This expression can be solved by algebraic substitution after making the following modifications:

$\int {\frac{dx}{x\cdot (x^{3}+1)} } = \int {\frac{3\cdot x^{2}}{3\cdot x^{3}\cdot (x^{3}+1)} } \, dx = \frac{1}{3}\int {\frac{3\cdot x^{2}}{x^{3}\cdot (x^{3}+1)} } \, dx$

And the following substitutions are used: u = x³, du = 3 · x² dx

$\frac{1}{3} \int {\frac{du}{u \cdot (u+1)} }$

This is expression is equivalent to the following by partial fraction decomposition:

$-\frac{1}{3}\int {\frac{du}{u+1} } \, du +\frac{1}{3}\int {\frac{du}{u} }$

And the integral is now solved:

$I = -\frac{1}{3}\cdot \ln |u+1| + \frac{1}{3}\cdot \ln |u| + C$

$I = \frac{1}{3}\cdot \ln \left|\frac{u}{u+1} \right| + C$

$I = \frac{1}{3}\cdot \ln \left|\frac{x^{3}}{x^{3}+1} \right| + C$

Where C is the integration constant.

After applying algebraic substitution, partial fraction decomposition and integration rules the integral of the non-trascendental expression $\frac{1}{x\cdot (x^{3}+1)}$ is equal to the trascendental expression $I = \frac{1}{3}\cdot \ln \left|\frac{x^{3}}{x^{3}+1} \right| + C$.

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