_ ^ x^2 dx (a + bx) ^2 = - Vidyadip

MCQ

$\int_{}^{} {\frac{{{x^2}dx}}{{{{(a + bx)}^2}}}} = $

A
$\frac{1}{{{b^2}}}\left[ {x + \frac{{2a}}{b}\log (a + bx) - \frac{{{a^2}}}{b}\frac{1}{{a + bx}}} \right]$
B
$\frac{1}{{{b^2}}}\left[ {x - \frac{{2a}}{b}\log (a + bx) + \frac{{{a^2}}}{b}\frac{1}{{a + bx}}} \right]$
C
$\frac{1}{{{b^2}}}\left[ {x + \frac{{2a}}{b}\log (a + bx) + \frac{{{a^2}}}{b}\frac{1}{{a + bx}}} \right]$
✓
$\frac{1}{{{b^2}}}\left[ {x + \frac{a}{b} - \frac{{2a}}{b}\log (a + bx) - \frac{{{a^2}}}{b}\frac{1}{{a + bx}}} \right]$

✓

Answer

Correct option: D.

$\frac{1}{{{b^2}}}\left[ {x + \frac{a}{b} - \frac{{2a}}{b}\log (a + bx) - \frac{{{a^2}}}{b}\frac{1}{{a + bx}}} \right]$

d
(d) Put $a + bx = t \Rightarrow x = \frac{{t - a}}{b}$ and $dx = \frac{{dt}}{b}$
$\therefore \,\,\,I = {\int_{}^{} {\left( {\frac{{t - a}}{b}} \right)} ^2} \times \frac{1}{{{t^2}}}\frac{{dt}}{b}$
$ = \frac{1}{{{b^2}}}\int_{}^{} {\left( {1 - \frac{{2a}}{t} + {a^2}.{t^{ - 2}}} \right)} \,dt = \frac{1}{{{b^2}}}\left[ {t - 2a\,\,\log t - \frac{{{a^2}}}{t}} \right]$
$ = \frac{1}{{{b^2}}}\left[ {x + \frac{a}{b} - \frac{{2a}}{b}\log (a + bx) - \frac{{{a^2}}}{b}\frac{1}{{(a + bx)}}} \right]$.

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