x , ,dx x^2 + 4x + 5 = - Vidyadip

MCQ

$\int {\frac{{x\,\,dx}}{{{x^2} + 4x + 5}} = } $

A
$\frac{1}{2}\log ({x^2} + 4x + 5) + 2{\tan ^{ - 1}}(x) + c$
B
$\frac{1}{2}\log ({x^2} + 4x + 5) - {\tan ^{ - 1}}(x + 2) + c$
C
$\frac{1}{2}\log ({x^2} + 4x + 5) + {\tan ^{ - 1}}(x + 2) + c$
✓
$\frac{1}{2}\log ({x^2} + 4x + 5) - 2{\tan ^{ - 1}}(x + 2) + c$

✓

Answer

Correct option: D.

$\frac{1}{2}\log ({x^2} + 4x + 5) - 2{\tan ^{ - 1}}(x + 2) + c$

d
(d)$I = \int {\frac{{x\,\,dx}}{{{x^2} + 4x + 5}}} $$ = \int {\frac{{x + 2 - 2\,\,}}{{{{(x + 2)}^2} + 1}}dx} $
$ = \frac{1}{2}\int {\frac{{2(x + 2)\,\,\,dx}}{{{{(x + 2)}^2} + 1}}} - 2\int {\frac{{dx}}{{1 + {{(x + 2)}^2}}}} $
$ = \frac{1}{2}\int {\frac{{dt}}{t} - 2\int {\frac{{dx}}{{1 + {{(x + 2)}^2}}}} } $

[Put $1 + {(x + 2)^2} = t$ in first expression ==> $2(x +2)dx = dt$]

$ = \frac{1}{2}\log t - 2{\tan ^{ - 1}}(x + 2) + c$
$ = \frac{1}{2}\log ({x^2} + 4x + 5) - 2{\tan ^{ - 1}}(x + 2) + c$.

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