x^2+3x+1 (x+1)^2 dx - Vidyadip

Question

$\int\frac{\text{x}^2+3\text{x}+1}{(\text{x}+1)^2}\text{dx}$

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Answer

$\int\Big(\frac{\text{x}^2+3\text{x}+1}{(\text{x}+1)^2}\Big)\text{dx}$
Let $\text{x}+1=\text{t}$
$\Rightarrow\text{x}=\text{t}-1$
$\Rightarrow1=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{dx}=\text{dt}$
Now, $\int\Big(\frac{\text{x}^2+3\text{x}+1}{(\text{x}+1)^2}\Big)\text{dx}$
$=\int\Big[\frac{(\text{t}-1)^2+3(\text{t}-1)+1}{\text{t}^2}\Big]\text{dt}$
$=\int\Big(\frac{\text{t}^2-2\text{t}+1+3\text{t}-3+1}{\text{t}^2}\Big)\text{dt}$
$=\int\Big(\frac{\text{t}^2+\text{t}-1}{\text{t}^2}\Big)\text{dt}$
$=\int\Big(1+\frac{1}{\text{t}}-\text{t}^{-2}\Big)\text{dt}$
$=\text{t}+\log|\text{t}|-\frac{\text{t}^{-2+1}}{-2+1}+\text{C}$
$=\text{t}+\log|\text{t}|-\frac{\text{1}}{\text{t}}+\text{C}$
$=\text{x}+1+\log|\text{x+1}|+\frac{1}{\text{x}+1}+\text{C}$
Let $1+\text{C}=\text{C}'$
$=\text{x}+\log|\text{x+1}|+\frac{1}{\text{x}+1}+\text{C}'$

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