Chapter G, Problem 33E

To determine

To calculate:

The integral value of the following:

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

Answer to Problem 33E

The integral value of ${\displaystyle \int \frac{x-3}{{\left(x^2+2x+4\right)}^2}dx}$ is $-\frac{2\arctan \left(\frac{x+1}{\sqrt{3}}\right)}{3^{\frac{3}{2}}}-\frac{4x+7}{6\left(x^2+2x+4\right)}+C$ .

Explanation of Solution

Given information:

The integral function:

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

Calculation:

The integral function is given that:

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

Now,

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

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

Simplify the first term of above right hand side of equation (1)

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

Now, substitute the value:

  $\begin{array}{l}u=x^2+2x+4\\ \frac{d}{dx}\left(u\right)=\frac{d}{dx}\left(x^2+2x+4\right)\\ \frac{du}{dx}=2x+2\\ dx=\frac{1}{2x+2}du\\ ={\displaystyle \int \frac{1}{2u^2}du}\end{array}$

Simplify the equation:

  ${\displaystyle \int \frac{x+1}{{\left(x^2+2x+4\right)}^2}dx}=\frac{1}{2}{\displaystyle \int \frac{1}{u^2}du}$

Know that:

  ${\displaystyle \int \frac{1}{u^2}du=-\frac{1}{u}\left[\because {\displaystyle \int u^{n}du=\frac{u^{n+1}}{n+1},n=-2}\right]}$

Put in solved integral:

  $\frac{1}{2}{\displaystyle \int \frac{1}{u^2}du=-\frac{1}{2u}}$

Substitute the value $u=x^2+2x+4$ :

  $\frac{1}{2}{\displaystyle \int \frac{1}{u^2}du=-\frac{1}{2\left(x^2+2x+4\right)}dx}$   ...(2)

Simplify the second term of above right hand side of equation (1)

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

  ${\displaystyle \int \frac{1}{{\left(x^2+2x+4\right)}^2}}dx={\displaystyle \int \frac{1}{{\left({\left(x+1\right)}^2+3\right)}^2}dx}\left[\because a^2+2ab+b^2={\left(a+b\right)}^2\right]$   ...(3)

Substitute:

  $\begin{array}{l}u=x+1\\ \frac{d}{dx}\left(u\right)=d\left(x+1\right)\\ \frac{du}{dx}=1\\ du=dx\end{array}$

The equation 3 rewritten as

  $\begin{array}{l}{\displaystyle \int \frac{1}{{\left(x^2+2x+4\right)}^2}}dx={\displaystyle \int \frac{1}{{\left(u^2+3\right)}^2}du}\\ \end{array}$

  ${\displaystyle \int \frac{1}{{\left(x^2+2x+4\right)}^2}}dx={\displaystyle \int \frac{u}{6\left(u^2+3\right)}du}+\frac{1}{6}{\displaystyle \int \frac{1}{u^2+3}du}$   ...(4)

By using this formula:

  ${\displaystyle \int \frac{1}{{\left(a{u}^2+b\right)}^n}du=\frac{2n-3}{2b\left(n-1\right)}{\displaystyle \int \frac{1}{{\left(a{u}^2+b\right)}^{n-1}}du+\frac{u}{2b\left(n-1\right){\left(au+b\right)}^{n-1}},a=1,b=3,n=2}}$ Simplify the first term of above right hand side of equation (4)

  ${\displaystyle \int \frac{u}{\left(u^2+3\right)}du}$

Substitute

  $\begin{array}{l}v=\frac{u}{\sqrt{3}}\\ \frac{d}{du}\left(v\right)=\frac{d}{du}\left(\frac{u}{\sqrt{3}}\right)\\ du=\sqrt{3}dv\end{array}$

  ${\displaystyle \int \frac{u}{\left(u^2+3\right)}du}=\frac{\sqrt{3}}{3v^2+3}dv$

Simplify the equation:

  ${\displaystyle \int \frac{u}{\left(u^2+3\right)}du}=\frac{1}{\sqrt{3}}{\displaystyle \int \frac{1}{v^2+1}dv}$  ...(5)

Know that, this is the standard integral.

  ${\displaystyle \int \frac{1}{v^2+1}dv}=\arctan \left(v\right)$

Put in solved integral:

  $\frac{1}{\sqrt{3}}{\displaystyle \int \frac{1}{v^2+1}dv}=\frac{\arctan \left(v\right)}{\sqrt{3}}$

Substitute the value $v=\frac{u}{\sqrt{3}}$

  $\frac{1}{\sqrt{3}}{\displaystyle \int \frac{1}{v^2+1}dv}=\frac{\arctan \left(\frac{u}{\sqrt{3}}\right)}{\sqrt{3}}$

Put the value in equation (4)

  ${\displaystyle \int \frac{1}{{\left(x^2+2x+4\right)}^2}}dx={\displaystyle \int \frac{u}{6\left(u^2+3\right)}du}+\frac{1}{6}{\displaystyle \int \frac{1}{u^2+3}du}$

  ${\displaystyle \int \frac{1}{{\left(x^2+2x+4\right)}^2}}dx={\displaystyle \int \frac{\arctan \left(\frac{u}{\sqrt{3}}\right)}{{2.3}^{\frac{3}{2}}}du}+\frac{u}{6\left(u^2+3\right)}$

Substitute $u=x+1$

  ${\displaystyle \int \frac{1}{{\left(x^2+2x+4\right)}^2}}dx={\displaystyle \int \frac{\arctan \left(\frac{x+1}{\sqrt{3}}\right)}{{2.3}^{\frac{3}{2}}}du}+\frac{x+1}{6\left({\left(x+1\right)}^2+3\right)}$

Put the solved integral in equation (1)

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

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

This problem is solved:

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

Simplify the equation:

  ${\displaystyle \int \frac{x-3}{{\left(x^2+2x+4\right)}^2}dx}=-\frac{2\arctan \left(\frac{x+1}{\sqrt{3}}\right)}{3^{\frac{3}{2}}}-\frac{4x+7}{6\left(x^2+2x+4\right)}+C$