: Let I = x 6 + 1 dx Put x 3 = t 3 x 2 dx = dt I = t 2 + 1 = - 1 t + C = - 1 ( x 3 ) + C

Integrals — Class 12 Maths Solution

ncert exercise SA NCERT,ex.7.4,Q.1,Page 315

Question

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

Step-by-step Solution

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

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

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

NCERT & Exemplar solution for CBSE Class 12 Mathematics, Integrals. Curated by Sachin Sharma. Free for all students.