: Let I = - 1 x 1 + x 2 dx Put - 1 x = t 1 + x 2 dx = dt I = dt = e t + C = e - 1 x + C

Integrals — Class 12 Maths Solution

ncert exercise SA NCERT,ex.7.2,Q.18,Page 305

Question

$\cfrac{{{e^{{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}$

Step-by-step Solution

: Let $I = \int {\cfrac{{{e^{{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}dx}$

Put ${\tan ^{ - 1}}x = t$ $\Rightarrow$ $\cfrac{1}{{1 + {x^2}}}dx = dt$
$\therefore$ $I = \int {{e^t}dt} = {e^t} + C = {e^{{{\tan }^{ - 1}}x}} + C$

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