Chapter G, Problem 21E

To determine

To calculate:

Integral of following function.

  ${\displaystyle \int \frac{5x^2+3x-2}{x^3+2x^2}dx}$ .

Answer to Problem 21E

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

Explanation of Solution

Given information:

  ${\displaystyle \int \frac{5x^2+3x-2}{x^3+2x^2}dx}$

Calculation:

Here given function is,

  ${\displaystyle \int \frac{5x^2+3x-2}{x^3+2x^2}dx}$

Now factor the denominator,

  $={\displaystyle \int \frac{5x^2+3x-2}{x^2\left(x+2\right)}dx}$

Partial fraction decomposition,

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

Now applying linearity,

  $=3{\displaystyle \int \frac{1}{x+2}dx+}2-{\displaystyle \int \frac{1}{x^2}dx+}$

Now solving first term,

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

Put up,

  $\begin{array}{l}u=x+2+\to \frac{du}{dx}=1\to dx=du:\\ ={\displaystyle \int \frac{1}{u}du}\end{array}$

This standard integral

  $=\ln \left(u\right)$

Undo the substation of the $u=x+2$

  $=\ln \left(x+2\right)$

Now solving second term,

  $={\displaystyle \int \frac{1}{x}dx}$

Make use of last results

  $=\ln \left(x\right)$

Now solving third term,

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

After applying power rule,

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

  $=-\frac{1}{x}$

Now plug all of the three integrals,

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