Question
Download Solution PDF\(\mathop \smallint \nolimits_1^2 \frac{1}{{{{\rm{x}}^2} + {\rm{x}}}}{\rm{dx}}\) किसके बराबर है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFअवधारणा:
\(\smallint \frac{1}{{\rm{x}}}{\rm{dx}} = \log \left| {\rm{x}} \right| + {\rm{c}}\)
गणना:
I = \(\mathop \smallint \nolimits_1^2 \frac{1}{{{{\rm{x}}^2} + {\rm{x}}}}{\rm{dx}} \)
\(= {\rm{\;}}\mathop \smallint \nolimits_1^2 \frac{1}{{{\rm{x}}\left( {{\rm{x}} + 1} \right)}}{\rm{dx}} \)
\(= \mathop \smallint \nolimits_1^2 \frac{{\left( {{\rm{x}} + 1} \right) - {\rm{x}}}}{{{\rm{x}}\left( {{\rm{x}} + 1} \right)}}{\rm{dx}} \)
\(= \mathop \smallint \nolimits_1^2 \left[ {\frac{{\left( {{\rm{x}} + 1} \right)}}{{{\rm{x}}\left( {{\rm{x}} + 1} \right)}} - \frac{{\rm{x}}}{{{\rm{x}}\left( {{\rm{x}} + 1} \right)}}} \right]{\rm{dx}} \)
\(= {\rm{\;}}\mathop \smallint \nolimits_1^2 \left[ {\frac{1}{{\rm{x}}} - \frac{1}{{{\rm{x}} + 1}}} \right]{\rm{dx}}\)
\( = {\rm{\;}}\left[ {\log x - \log \left( {x + 1} \right)} \right]_1^2\)
= (log 2 – log 3) – (log 1 – log 2)
= log 2 – log 3 – 0 + log 2
= 2 log 2 – log 3
= log 22 – log 3
(∵ n log m = log mn)
= log 4 – log 3
\( = \log \frac{4}{3}\)
(∵ log m - log n = log (m/n)
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