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

Question

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

✓

Answer

Let
$
\begin{aligned}
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
\end{aligned}
$
On comparing coefficients of $x$ and constants, Now
$
\begin{aligned}
2 A & =1 & \therefore A=\frac{1}{2} \\
I & =4 A+B & \\
I & =4 \times \frac{1}{2}+B & \therefore B=1-2=-1
\end{aligned}
$
now$\quad x+1=\frac{1}{2}(2 x+4)-1$
$
\begin{array}{l}
\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
\end{array}
$
For $I _1$ :
Let
$
\begin{aligned}
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
\end{aligned}
$
For $I _2$ :
$
\int \frac{1}{x^2+4 x+5} d x
$
Here
$
\begin{aligned}
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
\end{aligned}
$
Let $\quad x+2=t$ then $d x=d t$
$
\begin{aligned}
I_2 & =\int \frac{d t}{t^2+(1)^2}=\tan ^{-1} t+C_2 \\
& =\tan ^{-1}(x+2)+C_2
\end{aligned}
$
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 $C _2$ $I =\frac{1}{2} \log \left|x^2+4 x+5\right|-\tan ^{-1}(x+2)+C$ Ans.

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