Choose the correct answer in exercise: dx x(x^2+1) equals

MCQ

Choose the correct answer in exercise:
$\int\frac{\text{dx}}{\text{x}(\text{x}^2+1)}$ equals

✓
$\text{log}|\text{x}|-\frac{1}{2}\text{log}(\text{x}^2+1)+\text{C}$
B
$\text{log}|\text{x}|+\frac{1}{2}\text{log}(\text{x}^2+1)+\text{C}$
C
$-\text{log}|\text{x}|+\frac{1}{2}\text{log}(\text{x}^2+1)+\text{C}$
D
$\text{log}|\text{x}|+\frac{1}{2}\text{log}(\text{x}^2+1)+\text{C}$

✓

Answer

Correct option: A.

$\text{log}|\text{x}|-\frac{1}{2}\text{log}(\text{x}^2+1)+\text{C}$

Let $\text{I}=\int\frac{1}{\text{x}(\text{x}^2+1)}\text{dx}=\int\frac{2\text{x}}{2\text{x}^2(\text{x}^2+1)}\text{dx}=\int\frac{2\text{x}}{2\text{x}^2(\text{x}^2+1)}\text{dx}\dots(\text{i})$
Putting $x^2 = t$
$\Rightarrow 2x\ dx = dt$
Putting this value in eq. $(i),$
$\text{I}=\int\frac{\text{dt}}{2\text{t}(\text{t}+1)}=\frac{1}{2}\int\frac{(\text{t}+1)-\text{t}}{\text{t}(\text{t}+1)}\text{dt}$
$\text{I}=\frac{1}{2}\int\Bigg(\frac{\text{t}+1}{\text{t}(\text{t}+1)}-\frac{\text{t}}{\text{t}(\text{t}+1)}\Bigg)\text{dt}=\frac{1}{2}\int\Bigg(\frac{1}{\text{t}}-\frac{1}{\text{t}+1}\Bigg)\text{dt}=\frac{1}{2}\Bigg[\int\frac{1}{\text{t}}\text{dt}-\int\frac{1}{\text{t}+1}\text{dt}\Bigg]$
$=\frac{1}{2}[\text{log}|\text{t}|-\text{log}|\text{t}+1|]+\text{c}$
$=\frac{1}{2}[\text{log}|\text{x}^2|-\text{log}(\text{x}^2+1)]+\text{c}$
$=\text{log}|\text{x}|-\frac{1}{2}\text{log}|\text{x}^2+1|+\text{c}$

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