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प्रश्न
\[\int\frac{1}{2 - 3x} + \frac{1}{\sqrt{3x - 2}} dx\]
योग
उत्तर
\[\int\left( \frac{1}{2 - 3x} + \frac{1}{\sqrt{3x - 2}} \right)dx\]
\[ = \int\frac{dx}{2 - 3x} + \int \left( 3x - 2 \right)^{- \frac{1}{2}} dx\]
\[ = \frac{\ln \left( 2 - 3x \right)}{- 3} + \left[ \frac{\left( 3x - 2 \right)^{- \frac{1}{2} + 1}}{3\left( - \frac{1}{2} + 1 \right)} \right] + C\]
\[ = \frac{\ln \left( 2 - 3x \right)}{- 3} + \frac{2}{3} \left( 3x - 2 \right)^\frac{1}{2} + C\]
\[ = - \frac{1}{3}\ln \left( 2 - 3x \right) + \frac{2}{3}\sqrt{3x - 2} + C\]
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