: Let I = x + x x )dx Put 1 + x = t x dx = dt I = x ( 1 + x ) dx = t dt = t + C = ( 1 + x ) + C

Integrals — Class 12 Maths Solution

ncert exercise SA NCERT,ex.7.2,Q.3,Page 304

Question

$\cfrac{1}{{x + x\log x}}$

Step-by-step Solution

: Let $I = \int {\left( {\cfrac{1}{{x + x\log x}}} \right)dx}$
Put $1 + \log x = t$ $\Rightarrow$ $\cfrac{1}{x}dx = dt$

$\therefore$ $I = \int {\cfrac{1}{{x\left( {1 + \log x} \right)}}dx} = \int {\cfrac{1}{t}dt} = \log t + C = \log \left( {1 + \log x} \right) + C$

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