class 12 maths integrals

$\cfrac{{3{x^2}}}{{{x^6} + 1}}$

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📘 Integrals NCERT,ex.7.4,Q.1,Page 315 SA

$\cfrac{{3{x^2}}}{{{x^6} + 1}}$

Official Solution

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: Let $I = \int {\cfrac{{3{x^2}}}{{{x^6} + 1}}} dx$

Put ${x^3} = t$ $\Rightarrow$ $3{x^2}dx = dt$

$\therefore$ $I = \int {\cfrac{{dt}}{{{t^2} + 1}}} = {\tan ^{ - 1}}t + C = {\tan ^{ - 1}}\left( {{x^3}} \right) + C$

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