`int (2x-5) / (x^2+2x + 2) dx` Find or evaluate the integral by completing the square

Saturday, May 8, 2010

$\displaystyle\int\frac{{{2}{x}-{5}}}{{{{x}}^{{2}}+{2}{x}+{2}}}{\left.{d}{x}\right.}$ Find or evaluate the integral by completing the square

We have to evaluate the integral: $\int \frac{2x-5}{x^2+2x+2}dx$

We can write the integral as:

$\int \frac{2x-5}{x^2+2x+2}dx=\int\frac{2x-5}{(x+1)^2+1}dx$

Let $\displaystyle{x}+{1}={t}$

So, $\displaystyle{\left.{d}{x}\right.}={\left.{d}{t}\right.}$

Now we can write the integral as:

$\int \frac{2x-5}{(x+1)^2+1}dx=\int \frac{2(t-1)-5}{t^2+1}dt$

$=\int \frac{2t-7}{t^2+1}dt$

$=\int \frac{2t}{t^2+1}dt-\int\frac{7}{t^2+1}dt$ --------------->(1)

Now we will first evaluate the integral $\int \frac{2t}{t^2+1}dt$

Let $\displaystyle{{t}}^{{2}}+{1}={u}$

So, $\displaystyle{2}{t}{\left.{d}{t}\right.}={d}{u}$

Hence we can write,

$\int \frac{2tdt}{t^2+1}=\int \frac{du}{u}$

$\displaystyle={\ln{{\left({u}\right)}}}$

$\displaystyle={\ln{{\left({{t}}^{{2}}+{1}\right)}}}$

Now we will evaluate the second integral : $\int \frac{7}{t^2+1}dt$

$\int \frac{7}{t^2+1}dt=7\int \frac{1}{t^2+1}dt$

$=7tan^{-1}(t)$

Substituting both these integral results in (1) we get,

$\int \frac{2x-5}{x^2+2x+2}dx=ln(t^2+1)-7tan^{-1}(t)+C$ where C is a constant

$=ln((x+1)^2+1)-7tan^{-1}(x+1)+C$

$=ln(x^2+2x+2)-7tan^{-1}(x+1)+C$

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