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