class 12 maths integrals

$\cfrac{x}{{9 - 4{x^2}}}$

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📘 Integrals NCERT,ex.7.2,Q.15,Page 304 SA

$\cfrac{x}{{9 - 4{x^2}}}$

Official Solution

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: Let$I = \int {\cfrac{x}{{9 - 4{x^2}}}dx}$
Put $9 - 4{x^2} = t$ $\Rightarrow$ $- 8x\,dx = dt$

$\therefore$ $I = - \cfrac{1}{8}\int {\cfrac{{dt}}{t}} = - \cfrac{1}{8}\log \left| t \right| + C = \cfrac{1}{8}\log \cfrac{1}{{\left| t \right|}} + C$

$= \cfrac{1}{8}\log \cfrac{1}{{\left| {9 - 4{x^2}} \right|}} + C$

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