A) \[\frac{1}{{{b}^{2}}}\left[ x+\frac{2a}{b}log(a+bx)-\frac{{{a}^{2}}}{b}\left( \frac{1}{a+bx} \right) \right]+c\]
B) \[\frac{1}{{{b}^{2}}}\left[ x-\frac{2a}{b}log(a+bx)+\frac{{{a}^{2}}}{b}\left( \frac{1}{a+bx} \right) \right]+c\]
C) \[\frac{1}{{{b}^{2}}}\left[ x+\frac{2a}{b}log(bx+a)+\frac{{{a}^{2}}}{b}.\frac{1}{a+bx} \right]+c\]
D) \[\frac{1}{{{b}^{2}}}\left[ x+\frac{a}{b}-\frac{2a}{b}\log (a+bx)-\frac{{{a}^{2}}}{b}\left( \frac{1}{a+bx} \right) \right]+c\]
Correct Answer: D
Solution :
Let \[\int_{{}}^{{}}{I=\frac{{{x}^{2}}dx}{{{(a+bx)}^{2}}}}\] Put \[a+bx=t\Rightarrow dx=\frac{dt}{b}\] \[\therefore \] \[I=\frac{1}{b}\int_{{}}^{{}}{\frac{{{\left( \frac{t-a}{b} \right)}^{2}}}{{{t}^{2}}}}dt\] \[=\frac{1}{{{b}^{3}}}\int_{{}}^{{}}{\frac{{{t}^{2}}+{{a}^{2}}-2at}{{{t}^{2}}}}dt\] \[=\frac{1}{{{b}^{3}}}\int_{{}}^{{}}{\left( 1+\frac{{{a}^{2}}}{{{t}^{2}}}-\frac{2a}{t} \right)}\,dt\] \[=\frac{1}{{{b}^{3}}}\int_{{}}^{{}}{\left( t-\frac{{{a}^{2}}}{t}-2a\log t \right)}\,+c\] \[=\frac{1}{{{b}^{3}}}\left( a+bx-\frac{{{a}^{2}}}{a+bx}-2a\log (a+bx) \right)+c\] \[=\frac{1}{{{b}^{2}}}\left( x+\frac{a}{b}-\frac{{{a}^{2}}}{b}\left( \frac{1}{a+bx} \right)-\frac{2a}{b}\log (a+bx) \right)+c\]
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