\[\int_{{}}^{{}}{\frac{{{x}^{2}}dx}{{{(a+bx)}^{2}}}}=\] [IIT 1979]

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

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

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

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

\[\int_{{}}^{{}}{\frac{{{x}^{2}}dx}{{{(a+bx)}^{2}}}}=\] �

1.	\[\int_{{}}^{{}}{\frac{{{x}^{2}}dx}{{{(a+bx)}^{2}}}}=\] [IIT 1979]
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.	\[\frac{1}{{{b}^{2}}}\left[ x+\frac{2a}{b}\log (a+bx)+\frac{{{a}^{2}}}{b}\frac{1}{a+bx} \right]\]
D.	\[\frac{1}{{{b}^{2}}}\left[ x+\frac{a}{b}-\frac{2a}{b}\log (a+bx)-\frac{{{a}^{2}}}{b}\frac{1}{a+bx} \right]\]
Answer» E.

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\[\int_{{}}^{{}}{\frac{{{x}^{2}}dx}{{{(a+bx)}^{2}}}}=\] [IIT 1979]

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