Chapter G, Problem 30E

To determine

To calculate:

Integral of following function.

  ${\displaystyle \int \frac{x^3}{x^3+1}dx}$ .

Answer to Problem 30E

Integral of given function is $\frac{\ln \left(\left|x+1\right|\right)}{3}-\frac{\ln \left(x^2-x+1\right)}{6}-\frac{\arctan \left(\frac{2x-1}{\sqrt{3}}\right)}{\sqrt{3}}+x+C$ .

Explanation of Solution

Given information:

  ${\displaystyle \int \frac{x^3}{x^3+1}dx}$

Calculation:

Here given function is,

  ${\displaystyle \int \frac{x^3}{x^3+1}dx}$

Write $x^2$ as a $x^2+1-1$ and then split:

  $\begin{array}{l}={\displaystyle \int \left(\frac{x^3+1}{x^3+1}+\frac{1}{x^3+1}\right)dx}\\ ={\displaystyle \int \left(1-\frac{1}{x^3+1}\right)}dx\end{array}$

Apply integrity:

  $={\displaystyle \int 1dx}-{\displaystyle \int \frac{1}{x^2+1}dx}$

After solving,

  ${\displaystyle \int 1dx}$

Now apply constant rule,

  $=x$

  ${\displaystyle \int \frac{1}{x^2+1}dx}$

Factor the denominator,

  ${\displaystyle \int \frac{1}{\left(x+1\right)\left(x^2-x+1\right)}dx}$

Perform partial fraction decomposition,

  ${\displaystyle \int \frac{1}{3\left(x+1\right)}-\frac{x-2}{3\left(x^2-x+1\right)}dx}$

Apply integrity:

  $\begin{array}{l}\frac{1}{3}{\displaystyle \int \frac{1}{\left(x+1\right)}dx-\frac{1}{3}\frac{x-2}{\left(x^2-x+1\right)}dx}\\ {\displaystyle \int \frac{1}{\left(x+1\right)}dx}\end{array}$

Substitute the $u=x+1\to \frac{du}{dx}=1\to dx=du$

  $={\displaystyle \int \frac{1}{u}du}$

This is standard integral:

  $\begin{array}{l}u=x+1\\ =\ln \left(x+1\right)\end{array}$

Write $x-2$ as $\frac{1}{2}\left(2x-1\right)-\frac{3}{2}$ and the split.

  $\begin{array}{l}\frac{1}{2}{\displaystyle \int \frac{2x-1}{x^2-x+1}dx-\frac{3}{2}{\displaystyle \int \frac{1}{x^2-x+1}dx}}\\ ={\displaystyle \int \frac{1}{u}du}\end{array}$

Use the previous result $\ln \left(u\right)$

Undo substitution $u=x^2-x+1$

  $=\ln \left(x^2-x+1\right)$

And now solving

  $={\displaystyle \int \frac{1}{x^2-x+1}dx}$

Complete the square,

  $={\displaystyle \int \frac{1}{\left(x-\frac{1}{2}\right)+\frac{3}{4}}dx}$

Substitute $u=\frac{2x-1}{\sqrt{3}}\to \frac{du}{dx}=\frac{2}{\sqrt{3}}\to dx=\frac{\sqrt{3}}{2}du$

  $\begin{array}{l}={\displaystyle \int \frac{2\sqrt{3}}{3u^2+3}du}\\ =\frac{2}{\sqrt{3}}{\displaystyle \int \frac{1}{u^2+1}du}\end{array}$

This is standard integral,

  $=\arctan \left(u\right)$

Plug solved integrals in first terms,

  $\begin{array}{l}\frac{2}{\sqrt{3}}{\displaystyle \int \frac{1}{u^2+1}du}\\ =\frac{2\arctan \left(u\right)}{\sqrt{3}}\end{array}$

Plug solved integrals in second terms,

  $\begin{array}{l}\frac{1}{2}{\displaystyle \int \frac{2x-1}{x^2-x+1}du}-\frac{3}{2}{\displaystyle \int \frac{1}{x^2-x+1}dx}\\ =\frac{\ln \left(x^2-x+1\right)}{2}-\sqrt{3}\arctan \left(\frac{2x-1}{\sqrt{3}}\right)\end{array}$

Plug solved integrals in terms,

  $\begin{array}{l}\frac{1}{3}{\displaystyle \int \frac{1}{x+1}dx}-\frac{1}{3}{\displaystyle \int \frac{x-2}{x^2-x+1}dx}\\ =\frac{\ln \left(x^2-x+1\right)}{6}+\frac{\arctan \left(\frac{2x-1}{\sqrt{3}}\right)}{\sqrt{3}}+\frac{\ln \left(x+1\right)}{3}\end{array}$

Plug solved integrals,

  $\begin{array}{l}{\displaystyle \int 1dx-{\displaystyle \int \frac{1}{x^3+1}dx}}\\ =\frac{\ln \left(x^2-x+1\right)}{6}-\frac{\arctan \left(\frac{2x-1}{\sqrt{3}}\right)}{\sqrt{3}}-\frac{\ln \left(x+1\right)}{3}+x\\ =\frac{\ln \left(\left|x+1\right|\right)}{3}-\frac{\ln \left(x^2-x+1\right)}{6}-\frac{\arctan \left(\frac{2x-1}{\sqrt{3}}\right)}{\sqrt{3}}+x+C\end{array}$