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

Integrals — Class 12 Maths Solution

ncert exercise SA NCERT,ex.7.2,Q.35,Page 305

Question

$\cfrac{{{{\left( {1 + \log x} \right)}^2}}}{x}$

Step-by-step Solution

: Let $I = \int {\cfrac{{{{\left( {1 + \log x} \right)}^2}}}{x}} dx$

Put $1 + \log x = t$ $\Rightarrow$ $\cfrac{1}{x}dx = dt$

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

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