Evaluate x+1 x^2+4 x+5 d x. - Vidyadip

Question

Evaluate $\int \frac{x+1}{x^2+4 x+5} d x$.

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Answer

Let
$I =\int \frac{x+1}{x^2+4 x+5} d x$
$x+1 =A \frac{d}{d x}\left(x^2+4 x+5\right)+B$
$x+1 =A(2 x+4)+B$
On comparing coefficients of $x$ and constants, Now
$2 A =1 $
$\therefore A=\frac{1}{2}$
$I =4 A+B $
$I =4 \times \frac{1}{2}+B $
$\therefore B=1-2=-1$
now$ x+1=\frac{1}{2}(2 x+4)-1$
$\therefore \int \frac{x+1}{x^2+4 x+5} d x$
$=\int \frac{\frac{1}{2}(2 x+4)-1}{x^2+4 x+5} d x$
$=\frac{1}{2} \int \frac{2 x+4}{x^2+4 x+5} d x-\int \frac{1}{x^2+4 x+5} d x$
For $I _1$ :
Let
$x^2+4 x+5 =t$
$(2 x+4) d x =d t$
$I_1 =\frac{1}{2} \int \frac{d t}{t}$
$=\frac{1}{2} \log |t|+C_1$
$ =\frac{1}{2} \log \left|x^2+4 x+5\right|+C_1$
For $I _2$ :
$\int \frac{1}{x^2+4 x+5} d x$
Here
$x^2+4 x+5$
$ =x^2+4 x+2^2-2^2+5$
$ =(x+2)^2+1$
$ =\int \frac{1}{(x+2)^2+1} d x$
$=\int \frac{1}{(x+2)^2+(1)^2} d x$
Let $ x+2=t$ then $d x=d t$
$I_2 =\int \frac{d t}{t^2+(1)^2}=\tan ^{-1} t+C_2$
$ =\tan ^{-1}(x+2)+C_2$
Putting values of $I _1$ and $I _2$
$I=\left\{\frac{1}{2} \log \left|x^2+4 x+5\right|+C_1\right\}-\left\{\tan ^{-1}(x+2)+C_2\right\}$
On writing new constant $C$ in place of $C _1$ and $C _2$
$I =\frac{1}{2} \log \left|x^2+4 x+5\right|-\tan ^{-1}(x+2)+C$

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