Let I = 0 1 dx Let's put 1 + x 2 = t 2 2xdx = 2tdt xdx = tdt therefore, I = 1 2 t = [t] 1 2 = 2 - 1

Integrals — Class 12 Maths Solution

exemplar sa SA NCERT Exemp. Q. 32,Page 165

Question

$\int_0^1 {\frac{x}{{\sqrt {1 + {x^2}} }}} dx$

Step-by-step Solution

Let $I = \int_0^1 {\frac{x}{{\sqrt {1 + {x^2}} }}} dx$

Let's put $1 + {x^2} = {t^2}$

$\Rightarrow$ $2xdx = 2tdt$

$\Rightarrow$ $xdx = tdt$

therefore,$I = \int_1^{\sqrt 2 } {\frac{{tdt}}{t}}$

$= [t]_1^{\sqrt 2 } = \sqrt 2 - 1$

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